CBSE Important Questions Class 9 Maths Chapter 6: Measuring Space Perimeter and Area

Important Questions Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area

This page gives Important Questions Class 9 Maths Chapter 6 with full answers. The chapter is about finding perimeter and area. The 12 questions here cover shaded regions made of squares, quarter circles and semicircles, triangles on the same base, the median of a triangle, the area between two circles and rectangles made from smaller rectangles.

Every answer is solved step by step in simple language. A figure is given with each question, so you can see which part is shaded and which lengths are used. First try each question yourself, then check your working with the answer.

Q1. Find the area of the shaded region.

Eight squares of side 8 cm with shaded regions 1 to 8: four full squares, two quarter circles and two square-minus-quarter-circle regions

Answer: Side of each square = 8 cm

Radius of each quarter-circle = 8 cm.

Observe that:

Area of shaded region = Area of region 1 + Area of region 2 + Area of region 3 + Area of region 4 + Area of region 5 + Area of region 6 + Area of region 7 + Area of region 8

The total area of the four shaded squares (Region 1 + 2 + 3 + 4) = 4 × 82 = 256 cm2

Area of two quadrants (Region 5 + 6) = 2 × (πr2/4) = 2 × (π × 8 × 8)/4 = 32π

Area of (Region 7 + 8) = 2 × (Area of square − Area of quadrant)

= 2 × (64 − 16π)

Total area of the shaded region = 256 + 32π + 128 − 32π

= 384 cm2

Q2. In △ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ ∥ PD. PQ is joined (Fig. 6.34). Prove that Area (△BPQ) = (1/2) Area (△ABC).

Triangle ABC with D the midpoint of AB, P on BC, Q on AB with CQ parallel to PD, and triangle BPQ shaded

Answer: Given:

D is the midpoint of AB.

P lies on BC.

Q lies on AB such that:

CQ ∥ PD

We have to prove:

Area(â–³BPQ) = (1/2) Area(â–³ABC)

Step 1: Compare triangles â–³DPQ and â–³DPC

Triangles â–³DPQ and â–³DPC have the same base DP.

Also,

CQ ∥ DP

Therefore, the vertices Q and C lie on a line parallel to DP.

Hence, triangles â–³DPQ and â–³DPC are on the same base and between the same parallels.

Therefore,

Area(â–³DPQ) = Area(â–³DPC)

Step 2: Compare triangles â–³BPQ and â–³BDC

Since D lies on BQ,

Area(â–³BPQ) = Area(â–³BDP) + Area(â–³DPQ)

Also, since P lies on BC,

Area(â–³BDC) = Area(â–³BDP) + Area(â–³DPC)

But,

Area(â–³DPQ) = Area(â–³DPC)

Therefore,

Area(â–³BPQ) = Area(â–³BDC)

Step 3: Use the fact that D is the midpoint of AB

Since D is the midpoint of AB,

BD = DA

Triangles â–³BDC and â–³ADC have equal bases BD and DA and the same perpendicular height from C to line AB.

Therefore,

Area(â–³BDC) = Area(â–³ADC)

Hence,

Area(â–³BDC) = (1/2) Area(â–³ABC)

Since,

Area(â–³BPQ) = Area(â–³BDC)

we get,

Area(â–³BPQ) = (1/2) Area(â–³ABC)

Q3. In △ABC, AD is a median. If Area(△ABD) = 3x + 5 and Area(△ACD) = 5x − 13, find x and the area of △ABC.

Triangle ABC with median AD; the two parts have areas 3x + 5 and 5x − 13

Answer: Since AD is a median of â–³ABC, it divides the triangle into two triangles of equal area.

Therefore,

Area(â–³ABD) = Area(â–³ACD)

So,

3x + 5 = 5x − 13

18 = 2x

x = 9

Now,

Area(â–³ABD) = 3(9) + 5 = 32

and

Area(△ACD) = 5(9) − 13 = 32

Hence,

Area(â–³ABC) = 32 + 32 = 64

Therefore, x = 9 and Area(â–³ABC) = 64 sq. units

Q4. In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal.

Quarter circle with centre O and radius r, triangle AOB shaded, semicircle on AB with the crescent between the two arcs shaded and the unshaded part x between them

Answer: Let the area between the two shaded regions = x

Let OA = OB = r.

Then AOB is a right-angled triangle.

Also, AB2 = OA2 + OB2

AB2 = r2 + r2

AB2 = 2r2

So, AB = r√2

Now, AB is the diameter of the semicircle.

Radius of semicircle:

= AB/2

= r√2/2

Area of semicircle:

= (1/2) × π × (r√2/2)2

= (1/2) × π × (r2/2)

= πr2/4

Area of circular shaded region = πr2/4 − x

Area of quarter circle with centre O and radius r:

= (1/4) × πr2

= πr2/4

Area of triangular shaded region = πr2/4 − x

Therefore, the areas of the two shaded regions are equal.

Hence proved.

Q5. In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are A, B, C, as marked. Show that the area of the rectangle is 2(A + C)(B + C)/C.

Rectangle with three triangles of areas A, B and C; triangle C is right-angled with vertical side p and horizontal side q; distances x and y marked

Answer: Let the vertical side of triangle C be p.

Let the horizontal side of triangle C be q.

So, C = (1/2) × p × q

Therefore, pq = 2C

Let the horizontal distance from the left side of the rectangle to p be x.

Let the vertical distance from the bottom side of the rectangle to q be y.

Then, A = (1/2) × p × x

So, px = 2A

Also, B = (1/2) × q × y

So, qy = 2B

Now, the rectangle has:

Length = x + q

Breadth = y + p

Area of rectangle = (x + q)(y + p)

= xy + xp + qy + qp

Now, xp = 2A, qy = 2B and qp = 2C

Also, xy = (xp × qy)/pq

= (2A × 2B)/(2C)

= 2AB/C

Therefore, Area of rectangle

= 2AB/C + 2A + 2B + 2C

Taking 2/C common:

= 2(AB + AC + BC + C2)/C

Factorising:

= 2(A + C)(B + C)/C

Hence proved.

Q6. Fig. 6.53 shows two circles passing through each other's centres. Find the area of the region enclosed by the two circles in terms of the common radius r.

Two equal circles with centres A and B passing through each other's centres, meeting at C and D, common region shaded

Answer: The two circles are congruent and each passes through the centre of the other.

Let the common radius be r.

The common enclosed region is made of two equal circular segments.

Join the two centres and the two intersection points.

Each circle forms a sector of angle 120°.

Area of one sector:

= (120/360) × πr2

= πr2/3

The triangle ACD is isosceles with AC = AD = r.

In triangle AMC,

AM = r/2 [Since AB = r]

By Pythagoras theorem,

AC2 = AM2 + MC2

⇒ r2 = (r/2)2 + MC2

⇒ MC2 = 3r2/4

⇒ MC = √3r/2

⇒ CD = √3r [Since CD = 2MC]

Now, area of triangle ADC = (1/2) × CD × AM

= (1/2) × √3r × (r/2)

= √3r2/4

So, area of one segment = πr2/3 − √3r2/4

Area of common enclosed region:

= 2 × (πr2/3 − √3r2/4)

= 2r2(π/3 − √3/4)

= r2(2π/3 − √3/2)

Therefore, the required area is r2(2π/3 − √3/2).

Q7. In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) = Area (C).

Right-angled triangle C with sides a, b and hypotenuse c, semicircles on all three sides forming crescent regions A and B

Answer: Let the sides of the right-angled triangle be a, b and c, where c is the hypotenuse.

Area of a semicircle with diameter d:

= (1/2) × π(d/2)2 = πd2/8

So, area of semicircle on side a = πa2/8

Area of semicircle on side b = πb2/8

Area of semicircle on side c = πc2/8

Since the triangle is right-angled:

a2 + b2 = c2

Multiplying both sides by π/8:

πa2/8 + πb2/8 = πc2/8

Therefore,

Area of semicircle on side a + Area of semicircle on side b = Area of semicircle on side c

From the figure, after removing the common overlapping parts, we get:

Area(A) + Area(B) = Area(C)

Hence proved.

Q8. In Fig. 6.51 we see two concentric circles with a common centre O. A chord BC of the larger circle is drawn, touching the smaller circle at A. The length of BC is l. Show that the area of the green region enclosed between the two circles is (1/4)Ï€l2.

Two concentric circles with centre O, green ring between them, chord BC of the larger circle touching the smaller circle at A

Answer: Let the radius of the larger circle be R.

Let the radius of the smaller circle be r.

Since BC touches the smaller circle at A, OA is perpendicular to BC.

Also, since O is the centre of the larger circle, the perpendicular from O to chord BC bisects the chord.

So, AB = AC = l/2

Now, in right triangle OAB:

OB2 = OA2 + AB2

Here, OB = R, OA = r and AB = l/2

So, R2 = r2 + (l/2)2

Therefore, R2 − r2 = l2/4

Area of green region

= area of larger circle − area of smaller circle

= πR2 − πr2

= π(R2 − r2)

Substitute R2 − r2 = l2/4:

Area of green region

= π × l2/4

= (1/4)Ï€l2

Hence proved.

Q9. In Fig. 6.50, four semicircles have been drawn within the given square whose side is 2 units. The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.

Square of side 2 units with four semicircles on its sides forming a blue 4-petalled flower; one half-petal marked as section 1

Answer: Side of square = 2 units

Each semicircle has diameter = 2 units.

So, radius = 1 unit.

Each petal is formed by two quarter-circular arcs of radius 1.

So, perimeter of one petal

= 2 × length of one quarter-circle

= 2 × ((1/4) × 2πr)

= 2 × (π/2)

= π

There are 4 petals.

Total perimeter of flower

= 4 × π

= 4Ï€ units

Using π = 22/7:

Perimeter = 4 × 22/7 = 88/7 units

Area of one-half petal:

Petal (Section 1) is formed by the quarter of the circle.

The one quarter sector has area:

= (1/4) × π × 12

= π/4

The triangle formed (marked in the figure) between the two radii is a right triangle with legs 1 and 1.

Area of triangle = (1/2) × 1 × 1 = 1/2

So, area of section 1:

= π/4 − 1/2

There are all together 8 sections, so the total area of petals = 8(π/4 − 1/2)

= 2π − 4

Using π = 22/7:

Area = 2 × 22/7 − 4

= 44/7 − 28/7

= 16/7 square units

Therefore:

Perimeter of flower = 88/7 units

Area of flower = 16/7 square units

Q10. Show that the areas of the shaded blue triangle and the shaded red triangle are equal. Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.

Triangle ABC with BC trisected at D and E; triangle ABD shaded blue and triangle AEC shaded red

Answer: In the figure, lines are drawn from one vertex of a triangle to the points of trisection of the opposite side.

So, the opposite side is divided into three equal parts.

Hence, BD = DE = EC .................(1)

Both triangles have the same height, say h, because they share the same top vertex and their bases lie on the same straight line.

Area of blue triangle ABD = (1/2) × BD × h

Area of red triangle AEC = (1/2) × EC × h

From (1), BD = EC

So, ar(ABD) = ar(AEC)

Hence, area of blue triangle = Area of red triangle

Step 2: How to rearrange the blue triangle to cover the red triangle

Although the two triangles have different shapes, the blue triangle can be rearranged to form the red triangle.

Horizontal Slicing: Imagine cutting the blue triangle into many thin horizontal strips.

Shifting the Strips: Each strip can be moved sideways without changing its length or area.

Rearranging: Move the strips gradually to the right. The strips still have the same total area, but together they now form the shape of the red triangle.

Conclusion: Since no part is added or removed while rearranging the strips, the area remains unchanged.

Therefore, the blue triangle and the red triangle have equal areas.

Q11. The figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions A and B. Show that A and B have equal area.

Square PQRS with a quarter circle centred at S and semicircles on sides PS and SR, shaded regions A and B and unshaded regions C and D

Answer: Let there be four area sections A, B, C and D under the quarter arc.

Let the side of the square be 2r.

Then the radius of each semicircle is r.

Area of semicircle with base PS:

C + A = (1/2)Ï€r2

Area of semicircle with base SR:

D + A = (1/2)Ï€r2

The total area of two semicircles:

C + D + 2A = πr2 ……(1)

The quarter circle has radius 2r and it contains A, B, C and D sections of area.

Area of quarter circle:

A + B + C + D = (1/4) × π(2r)2

⇒ A + B + C + D = (1/4) × 4πr2

⇒ A + B + C + D = πr2 ……(2)

Since the right-hand sides of (1) and (2) are equal, we have

C + D + 2A = A + B + C + D

⇒ A = B

Therefore, Area of A = Area of B

Q12. The figure shows nine identical rectangles fitted together to make a larger rectangle whose area is 72 cm2. Find the perimeter of each small rectangle.

Large rectangle made of nine identical rectangles: four lying lengthwise in the top row and five standing in the bottom row

Answer: Let the length of each small rectangle be l cm.

Let the breadth of each small rectangle be b cm.

From the figure:

The top row has 4 rectangles placed lengthwise.

The bottom row has 5 rectangles placed breadth-wise.

So, the total widths are equal: 4l = 5b

Therefore, b = 4l/5

Height of large rectangle = l + b

Width of large rectangle = 4l

Area of large rectangle = 72 cm2

So, 4l(l + b) = 72

Substitute b = 4l/5:

4l(l + 4l/5) = 72

⇒ 4l(9l/5) = 72

⇒ 36l2/5 = 72

⇒ l2 = 10

⇒ l = √10

Now, b = 4√10/5 = 0.8√10

Perimeter of each small rectangle:

= 2(l + b)

= 2(√10 + 0.8√10)

= 2(1.8√10)

= 3.6√10 cm

Therefore, the perimeter of each small rectangle is 3.6√10 cm.

Related Study Material on Extramarks

Q.1 In a PQR, if 2P = 3Q = 2R, find the measures of P, Q and R.

Marks:4
Ans

We know that the sum of angles of a triangle is 180. P + Q + R = 180o …iGiven, 2 P = 3 Q=2 R P=32 QAnd, R = 32 QBy putting the values of P and R in i,we get 32 Q+ Q + 32 Q = 18082 Q=180 Q=3608=45 P = 32 Q=32—45=67.5And, R = 32 Q=32—45=67.5Hence, the measures of P , Q and R are 67.5,45 and 67.5 respectively.

Q.2


In the given figure, prove that exterior angle QSR = interior angles (SQP + QPR + PRS).

Marks:3
Ans


Join P and S and produce PS to T
TSQ = SQP + SPQ(i)
Similarly, TSR = SPR + SRP(ii)
By adding (i) and (ii)
(TSQ + TSR) = (SQP +SPQ + SPR + SRP)
As, TSQ + TSR = QSR and SPQ + SPR = QPR
QSR = (SQP + QPR +SRP)

Q.3


In the given figure, PQ
RS, SP PQ and PQS = 46o, then find the value of x and y.

Marks:2
Ans

Given, PQ
RS
As SQ is transversal
y = 46° [Alternate interior angles]
Again, as SP PQ
SPQ = 90°
In SPQ,
x + 90° + 46° = 180o [Sum of angles of a triangle is 180o]
x = 44°
Hence, in the given figure, x = 44° and y = 46°

Q.4 If one angle of a triangle is equal to the sum of other two, show that the triangle is right-angled-triangle.

Marks:1
Ans

Let x, y and z be three angles of a ABC.
We know that the sum of angles of a triangle is 180°
x + y + z = 180° (i)
As per the given condition
Let x = y + z
y + z + y + z = 180°
2(y + z) = 180°
y + z = 90°
x = 180° – 90° = 90° [From (i)]
ABC is a right-angled-triangle

Q.5


In the given figure AB
CD, find the values of x, y and z.

Marks:4
Ans


Given, AB
CD and A = 35 and C = 45
x = C = 45 [Alternate interior angles]
Similarly,
y = A = 35
Draw EF
AB
CD such that
z = 1 + 2
Since AB
EF and AF is the transversal
1 = A = 35
Similarly,
As, EF
CD and FC is transversal
2 = C = 45
z = 1 + 2 = 35 + 45 = 80
Hence, x = 45, y = 35 and z = 80.

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FAQs (Frequently Asked Questions)

The formula for arc length is l = 2πr × θ/360. Here r is the radius and θ is the central angle in degrees.

Ï€ is irrational because it cannot be written as a fraction a/b, where a and b are integers. Its decimal expansion has no repeating pattern.

Heron’s formula is A = √[s(s − a)(s − b)(s − c)]. Here s = (a + b + c)/2 for a triangle with sides a, b, c.

The formula for area of a sector is πr² × θ/360. Here r is radius and θ is the angle at the centre.

Brahmagupta’s formula is A = √[(s − a)(s − b)(s − c)(s − d)]. Here s = (a + b + c + d)/2.