Measuring Space Perimeter and Area Class 9 Extra Questions Maths Chapter 6

During revision, students quickly go through Extra Questions for Class 9 Maths and Ganita Manjari Class 9 Maths Chapter 6 Measuring Space Perimeter and Area Important Extra Questions and Answers for clarity.

Class 9 Measuring Space Perimeter and Area Extra Questions

Extra Questions of Measuring Space Perimeter and Area

Class 9 Maths Chapter 6 Extra Questions – Measuring Space Perimeter and Area Extra Questions Class 9

Measuring Space Perimeter and Area Class 9 Short Question Answer

Question 1.
The base of an isosceles triangle is 10 cm and one of its equal sides is 13 cm. Find its area using Heron’s formula.
Solution:
We have a = 10 cm, b = 13 cm, c = 13 cm
∴ s = \(\frac{a+b+c}{2}=\frac{10+13+13}{2}=\frac{36}{2}\) = 18 cm
Thus, area of A = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{18(18-10)(18-13)(18-13)}\)
= \(\sqrt{18 \times 8 \times 5 \times 5}=\sqrt{9 \times 2 \times 2 \times 4 \times 5 \times 5}\)
= 3 × 2 × 2 × 5
= 60 cm²

Question 2.
In the given figure, a square of diagonal 8 cm is inscribed in a circle. Find the area of shaded region.
Solution:
Let the side of the square be a cm.
So, radius of the circle, r = OA = \(\frac{AC}{2}\)
⇒ r = \(\frac{8}{2}\) = 4 cm

So, in right angled ∆ABC
AB² + BC² = AC²
⇒a² + a² = 8²
⇒ 2a² = 64
⇒ a² = \(\frac{64}{2}\)
⇒ a² = 32
Area of shaded part = Area of circle – Area of square
= πr² – a² = \(\frac{22}{7}\) × 4 × 4 – 32
= 16\(\left[\frac{22}{7}-\frac{2}{1}\right]\) = 16\(\left[\frac{22-14}{7}\right]\)
= \(\frac{16 \times 8}{7}=\frac{128}{7}\)
Area of shaded region = 18\(\frac{2}{7}\) cm²

Question 3.
In the given figure, arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid points D, E and F respectively. Find the area of the shaded region. (Use π = 3.14)

Solution:
From the given figure, area of the shaded part is equal to the sum of areas of three sectors at points A, B and C. As ∆ABC is equilateral triangle of side 10 cm and radius of the sector is half of the side. All the three sectors are identical.
Angle of the sector, θ = 60°
Radius of each sector (r) = \(\frac{10}{2}\) = 5 cm
∴ Area of shaded part = 3 × (Area of sector)
= \(\frac{1}{2}\)
= \(\frac{1}{2}\)
= 1.57 × 25 = 39.25 cm²
Hence, the required area is 39.25 cm².

Question 4.
In the given figure, calculate the area of shaded portion.

Solution:
In right ∆PSQ, by Baudhayana-Pythagoras theorem
PQ² = PS² + SQ² = (12)² + (16)²
= 144 + 256 = 400 cm²
⇒ PQ = \(\sqrt{400}\) cm = 20 cm
Now a = 20 cm, b = 48 cm and c = 52 cm
s = \(\frac{a+b+c}{2}=\frac{20+48+52}{2}\)
= \(\frac{120}{2}\)cm = 60 cm
Thus, area of ∆PQR = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{60(60-20)(60-48)(60-52)}\)
= \(\sqrt{60 \times 40 \times 12 \times 8}\)
= \(\sqrt{3 \times 20 \times 20 \times 2 \times 3 \times 4 \times 4 \times 2}\)
= 3 × 20 × 2 × 4
= 480 cm²
Area of right APSQ = \(\frac{1}{2}\) × QS × PS = \(\frac{1}{2}\) × 16 × 12 = 96 cm²
∴ Area of shaded portion in the given figure
= 480 cm² – 96 cm² = 384 cm².

Question 5.
Four decorative lights are fixed at four points on the boundary of a circular stage. The lengths of wires joining consecutive lights are 6 m, 8 m, 10 m, and 12 m. Find the area enclosed by the wires.
Solution:
Since all four points lie on a circle, the quadrilateral formed is cyclic.
Given, a = 6m, b = 8 m, c = 10 m, and d = 12 m.
Semi-perimeter, (s) = \(\frac{6+8+10+12}{2}\) = 18 m
Using Brahmagupta’s formula we have,
Area = \(\sqrt{(18-6)(18-8)(18-10)(18-12)}\)
= \(\sqrt{12 \times 10 \times 8 \times 6}=\sqrt{5760}\)
∴ Area ≈ 75.89 m²

Question 3.
Find the difference of the areas of a sector of angle 120° and its corresponding major sector of a circle of radius 21 cm.

Solution:
For minor sector
r= 21 cm, θ, = 120°
For major sector
r = 21 cm,
θ₂ = 360 – 120° = 240°
Difference in areas of major and minor sectors
= \(\frac{\pi r^2}{360^{\circ}}\)(θ₂ – θ₁) [∵ r₁ = r₂ = r]
= \(\frac{22 \times 21 \times 21}{7 \times 360^{\circ}}\)
= \(\frac{22 \times 21 \times 21 \times 120^{\circ}}{7 \times 360^{\circ}}\)
= 462 cm²
Hence, the difference in areas of major and minor sectors of given circle is 462 cm².

Question 4.
Calculate the area other than the area common between two quadrants of the circles of radius 16 cm each, which is shown as the shaded region in the figure.

Solution:
Area of sector ADB


∴ Area of the shaded region-I
[Area of the sector ADB] – [Area of the sector ADB]
= 256 cm² – \(\frac{1408}{7}\)cm²
= \(\frac{1792-1408}{7}\)cm²
= \(\frac{384}{7}\)cm²
Similarly, the area of the shaded region-II = \(\frac{384}{7}\)cm
Total area of the shaded region
= [Area of shaded region-I] + [Area of shaded region-II]
= \(\frac{384}{7}\)cm² + \(\frac{384}{7}\)cm²
= \(\frac{768}{7}\)cm

Question 5.
PQRS is a square land of side 28 m. Two semi-circular grass covered postions are to be made on two of its opposite sides as shown in the figure. How much area will be left uncovered? [Take π = \(\frac{22}{7}\)]

Solution:
Side of the square = 28 m
∴ Area of the square PQRS = 28 × 28 m²
Diameter of a semi-circle = 28 m
Radius of a semi-circle =14 m
:. Area of semi-circIe= \(\frac{1}{2}\)πr² = \(\frac{1}{2} \times \frac{22}{7}\) × 14 × 14
= 22 × 14 m²
= 308 cm²
⇒ Area of both the semi-circles = 2 × 308 m² = 616 m²
∴ Area of the square left uncovered= (28 × 28) – 616 m²
= 784 – 616 m²
= 168 m²

Question 6.
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use n = 3.14)
Solution:
Here, Radius (r) = 16.5 km
Sector angle (θ) = 80°
∴ Area of the sea surface over which the ships are warned

= 189.97 km²

Question 7.
A round table cover has six equal designs as shown in Fig. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm². (Use √3 = 1.73)
Solution:
Here, r = 28 cm
Since, the circle is divided into six equal sectors.
∴ Sector angle θ = \(\frac{1}{2}\) = 60°.
∴ Area of the sector with 0 = 60° and r = 28 cm
= \(\frac{60}{360} \times \frac{22}{7}\) × 28 × 28 cm²
= \(\frac{44 \times 28}{3}\) cm²
=410.67 cm² …(i)

Now, area of 1 design = Area of segment APB
= Area of sector – Area of ΔAOB …(ii)
In A AOB, ∠AOB = 60°, OA = OB = 28 cm
∴ ∠OAB = 60° and ∠OBA = 60°
⇒ ΔAOB is an equilateral triangle.
⇒ AB = AO = BO
⇒ AB = 28 cm
Area of ΔAOB = \(\frac{1}{2}\) × 28 × 28 4
= 196√3
= 196 × 1.73
= 339.08 cm² …(iii)
Now, from (i), (ii) and (iii), we have:
Area of segment APB = 410.67 cm² – 339.08 cm²
= 71.59 cm²
⇒ Area of 1 design = 71.59 cm²
∴ Area of the 6 equal designs = 6 × (71.59) cm²
= 429.54 cm²
Cost of making the design at the rate of ₹ 0.35 per cm²
= ₹ 0.35 × 429.54
= ₹ 150.339

Measuring Space Perimeter and Area Class 9 Long Question Answer

Question 1.
The perimeter of a right triangle is 144 cm and its hypotenuse measure 65 cm. Find the length of other sides and calculate its area. Verify the result using Heron’s formula.
Solution:
Let the other two sides be a cm and 6 cm
Then a² + b² = (65 )²
⇒ a² + b² = 4225
Also perimeter of triangle = 144 cm
⇒ a + b + 65 = 144 ⇒ a + b = 79
Squaring (ii), we get
(a + b)² = (79)²
⇒ a² + b² + 2ab = 6241
⇒ 4225 + 2ab = 6241 [Using (i)]
⇒ 2ab = 2016
Also (a – b)² = a² + b² – 2ab = 4225 – 2016 = 2209
⇒ a – b = \(\sqrt2209}\) = 47
Solving (ii) and (iii), we get a = 63 cm, b = 16 cm
∴ The other two sides arc 63 cm and 16 cm.
Area of triangle = \(\frac{1}{2}\) × b × a = \(\frac{1}{2}\) × 16 × 63 = 504 cm

Here a = 63 cm, b = 16cm, c = 65 cm.
s = \(\frac{63+16+65}{2}=\frac{1}{2}\) × 144 cm = 72 cm

Thus, Area of A = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{72(72-63)(72-16)(72-65)}\)
= \(\sqrt{72 \times 9 \times 56 \times 7}\)
= \(\sqrt{8 \times 9 \times 9 \times 8 \times 7 \times 7}\)
= 8 × 9 × 7
= 504 cm²

Question 2.
The area of an equilateral triangle is 49√3 cm². Taking cacti angular point as centre, circles are drawn with radius equal to half length of the side of the triangle- Find the area of triangle not included in the circles, [Take √3 = 1.73]
Solution:
Let the side of equilateral triangle be a.

Area of ∆ABC = \(\frac{\sqrt{3}}{4}\) × a²
49√3 = \(\frac{\sqrt{3}}{4}\)a²
a² = 49 × 4
a = 14 cm
Since each angle of an equilateral triangle = 60°
Area of a sector having θ as 60° and radius \(\frac{14}{2}\), 7 cm.
= \(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × 7 × 7 cm² = \(\frac{11 \times 7}{3}\)cm²
[Using area = \(\frac{\theta}{360^{\circ}}\) × πr²]
∴ Area of 3 sectors = 3\(\frac{1}{2}\)cm² = 77 cm²
∴ Area of the shaded region
= (Area of equilateral ∆ABC) – (Area of 3 sectors)
= 49√3 – 11
= 84.77 – 77
= 7.77 cm².

Measuring Space Perimeter and Area Class 9 Case Based Questions

Question 1.
A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat.
Designs of some brooch are shown below. Observe them carefully.

Design A: Brooch A is made with silver wire in the form of a circle with diameter 28 mm. The wire used for making 4 diameters which divide the circle into 8 equal parts. Design B: Brooch B is made two colours: Gold and silver. Outer part is made with Gold. The circumference of silver part is 44 mm and the gold part is 3 mm wide everywhere.
(i) Find the total length of silver wire required to make design A.
(ii) What is the area of each sector of the brooch in design A?
(iii) Find the area of golden part in the brooch in design B.
OR
A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80π mm?
Solution:
Diameter of the circle is 28 mm.
∴ Radius (r) = \(\frac{28}{2}\) = 14 mm
The total length of silver wire required to make design A
= 2πr + 4d
= 2 × \(\frac{22}{7}\) × 14 mm + 4 × 28 mm
= 88 + 112
= 200 mm

(ii) The area of each sector of the brooch in design A.
= \(\frac{1}{8}\)πr²
= \(\frac{1}{8} \times \frac{22}{7}\) × (14)²
= 77 mm²

(iii) Circumference = 2πr
So, for silver part in design B,
44 mm = 2 × \(\frac{22}{7}\) × r
⇒ r = 44 × \(\frac{7}{2 \times 22}\)
= 7 mm
So, for gold part in design B,
R = r + 3 = 10 mm

Hence, the difference of areas of golden and silver parts
= π(R² – r²) = π( 100 – 49)
= 51 π mm
OR
Circumference of outer circle
= 2πR = 2π × 10
= 20n mm
The number of complete revolutions must brooch B take to cover 80π mm
= \(\frac{80 \pi}{20 \pi}\) = 4

Measuring Space Perimeter and Area Class 9 Competency Based Questions

Question 1.
The area of trapezium ABCD is 352 cm². Length of side BC is 10 cm. The distance between parallel sides is 19 cm and one of the parallel sides is 19 cm, the perimeter of isosceles AEBC is

(a) 22 cm
(b) 23 cm
(c) 26 cm
(d) 24 cm
Solution:
Let the other parallel side be x cm.
∴ Area of trapezium = – (sum of ∥ sides) × height
⇒ 352 = \(\frac{1}{2}\) × (19 + x) × 16
⇒ 19 + x = \(\frac{352 \times 2}{16}\) = 44
⇒ x = 44 – 19 = 25 cm
∴ EB = AB – AE = 25 – 19 = 6 cm
So, perimeter of AEBC = EB + BC + CE = 6 + 10 + 10
= 26 cm.
Hence, (c) is the correct answer.

Question 2.
If every side of a triangle is doubled, then increase in area of the triangle is
(a) (√2 × 100)%
(b) 200%
(c) 300%
(d) 400%
Solution:
Let the three sides of the triangle be a, b, c and 5 be its semi-perimeter.
When every side of the triangle doubled, new sides becomes 2a, 2b, 2c.
So, we have s = \(\frac{a+b+c}{2}\)
a + b + c = 2s

Also, S = \(\frac{2 a+2 b+2 c}{2}\)
S = 2 × \(\frac{a+b+c}{2}\) = 2s
∴ S – 2a = 2s – 2a = 2(s – a)
Similarly S – 2b = 2(s – b) and S – 2c = 2(5 – c)
∴ Area of new triangle
= \(\sqrt{\mathrm{S}(\mathrm{~S}-2 a)(\mathrm{S}-2 b)(\mathrm{S}-2 c)}\)
= \(\sqrt{2 s \times 2(s-a) \times 2(s-b) \times 2(s-c)}\)
= 4\(\sqrt{s(s-a)(s-b)(s-c)}\)
= 4 × Area of original triangle = 4∆
So, increase in area of the triangle
= (\(\frac{3 \Delta}{\Delta}\) × 100)% = 300%
Hence, (c) is the correct answer.

Question 3.
In the figure below RT = 1 cm and OQ = 3 cm.

What is the area of the shaded region?
(a) (12.5π – 12) cm²
(b) (6.25π – 12) cm²
(c) (12.5π – 15) cm²
(d) (6.25π – 15) cm²
Solution:
In ∆OTS, using Baudhayana-Pythagoras theorem, we get
r² = 3² + (r – 1)²
r² = 9 + r² + 1 – 2r
2r = 10 ⇒ r = 5 cm

Let radius is r cm.
Area of shaded region
= Area of quadrant – area of rectangle ORSQ
= \(\frac{1}{4}\)(5)² – 4 × 3
= (6.25π – 12) cm²
Hence, (b) is the correct answer.

Question 4.
A regular pentagon is inscribed in a circle with centre O, of radius 5 cm, as shown below.

What is the area of the shaded part of the circle?
(a) 2π cm²
(b) 4π cm²
(c) 5π cm²
(d) 10π cm²
Solution:
θ = \(\frac{360^{\circ}}{5}\) = 72°
Area of shaded part = 2 × area of sector
= 2 × \(\frac{72^{\circ}}{360^{\circ}}\) × π × (5)2 = 10π cm²
Hence, (d) is the correct answer.

Question 5.
In figure, OPQR is a rhombus, three of whose vertices lie on circle with centre O. If the area of the rhombus is 32√3 cm², find the radius of the circle.

Solution:
Let the radius of the circle be r cm. OPQR is a rhombus.
∴ All sides of rhombus are equal.
Therefore, we have OP = PQ = QR = OR
Also OP = OQ [Each is radius of circle.]
Now OP = PQ = OQ = r cm
⇒ AOPQ is an equilateral triangle.
Now area of rhombus = 2(area of ∆OPQ) = 2\(\left(\frac{\sqrt{3}}{4} r^2\right)\) cm
[∵ Area of equilateral triangle = \(\frac{\sqrt{3}}{4}\) (side)²]
Area of rhombus = 32√3 cm
2\(\left(\frac{\sqrt{3}}{4} r^2\right)\) = 32√3
r² = 32√3 × \(\frac{4}{2 \sqrt{3}}\) = 64
∴ r = \(\sqrt{64}\) = 8
Hence, radius of the circle is 8 cm

Question 6.
In given figure ∆ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of same area as that of ΔABC is constructed. Find the height DF of the parallelogram.

Solution:
Sides of triangle ABC are 7.5 cm, 7cm and 6.5 cm.
The semi-perimeter of ∆ABC
s = \(\frac{7.5+7+6.5}{2}=\frac{21}{2}\) = 10.5 cm
Area of ∆ABC = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{10.5(10.5-7.5)(10.5-7)(10.5-6.5)}\)
= \(\sqrt{10.5 \times 3 \times 3.5 \times 4}\)
= \(\sqrt{441}\)
= 21 cm²
Now as on base BC a parallelogram DBCE of same area as that of AABC is constructed.
Therefore, area of || gm DBCE = 21 cm²
Also area of ||gm DBCE = BC × DF
∴ BC × DF = 21 cm²
⇒ 7 × DF = 21
⇒ DF = 21 +7 = 3 cm
Hence, the height DF of the parallelogram is 3 cm.

Question 7.
Avikant bought a pair of glasses with wiper blades. He was curious to know the area being cleaned by each of the wiper blades. With the help of a ruler and a protractor, he found the length of each blade as 3 cm and the angle swept as 60°.
(Note: The figure is for visual representation only.)

(i) Find the area that each wiper cleans in one swipe, in terms of π.
(ii) If the diameter of each circular glass is 5 cm, what percent of the area of the glass will be cleaned by the blade in one swipe?
Solution:
(i) Area that each wiper cleans in one swipe
= \(\frac{60^{\circ}}{360^{\circ}}\) × π × (3)² = 1.5π cm²

(ii) Area of the glass as π × \(\frac{5}{2} \times \frac{5}{2}=\frac{25 \pi}{4}\) cm²
Percentage of the area cleaned by the wiper blade in one swipe as \(\frac{\frac{1.5 \pi}{25 \pi}}{4}\) × 100 = 24%.

Question 8.
A regular octagon of side length 4 cm is inscribed in a circle of radius 7 cm. A square is inscribed in the same circle as shown below.

Area of the shaded region. Show your work.
(Note: If needed, take π as
\(\frac{22}{7}\), √3 as 1.7, √5 as 2.2)
Solution:
The octagon divides the circles into 8 equal sectors of 45° each so, the area of each of the sectors as:
\(\frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × 7 × 7 = \(\frac{77}{4}\) = 19.25 cm2.
Here, a = 7 cm, 6 = 7 cm, c = 4 cm

Semi-perimeter of ∆VOU = \(\frac{1}{2}\) = 9 cm(= s)
Using Heron’s formula, Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)
Area of ∆VOU = \(\sqrt{(9 \times 2 \times 2 \times 5)}\) = 6√5 = 13.2 cm².
Area of the 8 segments = 8(19.25 – 13.2) = 48.4 cm².
Diagonal of square = 2r = 2 × 7 cm = 14 cm
Diagonal of square = √2 × side
Side = \(\frac{14}{\sqrt{2}}\)
Area of square PRTV = \(\frac{14}{\sqrt{2}} \times \frac{14}{\sqrt{2}}\) = = 98 cm².
Area of the shaded region = 48.4 + 98 = 146.4 cm².

Question 9.
Shown below are two circles with centres P and Q. Diameter ST is 6 cm.

Find the area of the shaded region. Draw a rough diagram and show your work.’
Solution:
Join ST such that it passes through centre Q.
Using the property, the angle subtended by a diameter at any point on the circle is 90°, then ∠SPT = 90°.
SP = PT
(Radii of big circle with centre P)
Using Baudhayana-Pythagorus theorem,
SP² + PT² = 6²
⇒ 2SP² = 36
⇒ SP² = 18

Area of ∆SPT = \(\frac{1}{2}\) × SP × PT
= \(\frac{1}{2}\) × SP²
= \(\frac{1}{2}\) × 18
= 9 cm²

Area of sector PST in circle with centre P
= \(\frac{1}{2}\) × π × 3²
= \(\frac{9 \pi}{2}\) cm²

Measuring Space Perimeter and Area Class 9 Extra Questions for Practice

Multiple Choice Questions

Question 1.
If the sides of a triangle are 56 cm. 60 cm and 52 cm long, then the area of the triangle is
(a) 1322 cm²
(b) 1311 cm²
(c) 1344cm²
(d) 1392cm²

Question 2.
If the area of a circle is 154 cm², then its perimeter is
(a) 11cm
(b) 22cm
(c) 44cm
(d) 55cm

Question 3.
The edges of a triangular board are 6cm, 8cm and 10cm. The cost of painting it at the rate of 9 paise per cm2 is
(a) ₹ 2.00
(b) ₹ 2.16
(c) ₹ 2.48
(d) ₹ 3.00

Assertion Reason Questions

Two statements are given, one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from
the options (a), (b), (c), and (d) given below.
(a) Both Assertion (A) and Reason (R) are tme and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A),
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.

Question 1.
Assertion (A): The area of a triangle is 24 cm² whose sides are 6 cm, 8 cm and 10 cm respectively.
Reason (R): Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)

Question 2.
Assertion (A):In triangle PQR. if arcs have been drawn with radii r cm each with centres P. Q and p R, then the area of the shaded region is \(\frac{1}{2}\)πr².
Reason (R): 1f the length of an arc of a circle of radius ris equal to that of an arc ola circle of radius 2r, then the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle.

Short Answer Type Questions

Question 1.
If the area of an equilateral triangle is 36√3 cm², find its height.

Question 2.
Three horses are tethered with 7 m long ropes at the three comers of a triangular field having sides 20 m, 34 m and 42 m. Find the area of the plot that can be grazed by the horses. Also, find the area of the plot which remains ungrazed.

Question 3.
The minute hand of a clock is 12 cm long. Find the area on the face of the clock described by the minute hand between 8:00 a.m. and 8:35 a.m.

Question 4.
Sides of a triangle are in the ratio 12 : 17 : 25 and its perimeter is 540 cm. Find its area.

Question 5.
The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the
first two sides is a right angle. Find its area.

Question 6.
In the figure, find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Long Answer Type Questions

Question 1.
The dimensions of a rectangle ABCD are 51 cm × 25 cm. A trapezium with its parallel sides QC and PD in the ratio 9 : 8, is cut off from the rectangle (see figure). If the area of the trapezium PQCD is T th part of the area of the rectangle, find the lengths QC and PD.

Question 2.
Find the area of the shaded region in the given figure, where ABCD is a square of side 28 cm.

The post Measuring Space Perimeter and Area Class 9 Extra Questions Maths Chapter 6 appeared first on Learn CBSE.

from Learn CBSE https://ift.tt/NVgTx7p
via IFTTT