 # Mensuration Area Questions Quiz 3 For SSC and Railways Exams with Solution

Important for :
1
ABC is a right angled triangle. B being the right angle. Mid-points of BC and AC are respectively B’ and A’. Area of ΔA’B’C is
» Explain it
C In ΔABC and A' B' C

A' B' || AB

∠B' = ∠B, ∠A' = ∠A

∴    Δ ABC ~ Δ A' B' C

 ∴     A' B' = 1 AB 2

(Ref: Mid point Theorem which states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.)

 ∴    Area of Δ A'B'C = 1 × B' C × A' B' 2

 = 1 × 1 × BC × 1 AB 2 2 2

 = 1 ( 1 × BC × AB) 4 2

 = 1 (Area of ΔABC) 4

Hence, option C is correct.

2
ABCD is parallelogram. P and Q are the mid-points of sides BC and CD respectively. If the area of ΔABC is 12 cm2, then the area of ΔAPQ is
» Explain it
C In the given ||gm P and Q are the mid points of the sides BC and CD respectively, and AC and BD are the diagonals.

In ΔABC, AP is a median and similarly in ΔADC, AQ is a median.

Now, since median divides the triangle into two equal areas, area of ΔABP = 1/2 area of ABC = 1/2 × 1/2 area of the ||gm ABCD

[Since the diagonal AC divides of the ||gm into two equal areas]

 = 1 of area of ||gm ABCD    ...(1) 4

 = 1 of area of ||gm ABCD   ...(2) 4

Now, in ΔBCD, if we join the mid-points P and Q, the area of the triangle thus formed (ΔPCQ) will be 1/4 of the area of the ΔBCD as per the mid-point theorem.

 ∴   Area of ΔPCQ = 1 of area of ΔBCD 4

 = 1 of 1 of area of ||gm ABCD 4 2

 = 1 of area of ||gm ABCD 8

∴  area of ΔAPQ

= (area ||gm ABCD) – [area ΔABP + area ΔADQ + area ΔPCQ]

= (area ||gm ABCD) –

 [( 1 + 1 + 1 ) of area of ||gm ABCD ] 4 4 8

= (area ||gm ABCD) – 5/8 × (area of ||gm ABCD)

 = 3 area ||gm ABCD 8

 = 3 (2 × area of ΔABCD) 8

 = 3 of area ΔABC 4

 = 3 × 12 = 9 cm2 4

Hence, option C is correct.

3
The difference between the radii of bigger circle and smaller circle is 14 cm and the difference between their areas is 1056 cm2. Radius of the smaller circle is
» Explain it
B
Radius of the larger circle = R cm

Radius of the smaller circle = r cm

∴    R – r = 14 cm

and π (R2 – r2) = 1056

 ⇒    R2 – r2 = 1056 = 1056 × 7 π 22

⇒ R2 – r2 = 336

⇒ (R – r)(R + r)= 336

⇒ (R + r) × 14 = 336

 ⇒ (R + r) = 336 = 24 cm 14

∴    (R + r) – (R – r) = 24 – 14

⇒    2r = 10   ⇒   r = 5 cm.

Hence, option B is correct.

4
In ΔPQR, the line drawn from the vertex P intersects QR at point S. If QR = 4.5 cm and SR = 1.5 cm then the ratios of the area of triangle PQS and triangle PSR is
» Explain it
D As seen in the diagram,

QR = 4.5 cm

SR = 1.5 cm

∴     QS = 4.5 – 1.5 = 3 cm

 Δ PQS Δ PSR
=
 1 × h × QS 2
 1 × h × SR 2

 = 3 = 2 : 1 1.5

Hence, option D is correct.

5
A circular wire of diameter 112 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 9 : 7. The smaller side of the rectangle is
» Explain it
A
Smart Approach

If the ratio of two sides of the gained rectangle is 9:7, the smaller side must be a multiple of 7. But if we closely look at the given options, none except option A; 77 is completely divisible by 7. Hence, we can immediately pick the option 'A' as our answer.

Circumference of circular shape = π × diameter

 = 22 × 112 = 352 cm 7

Length of wire = 352 cm

∴    Perimeter of rectangle = 2(length + breadth) ⇒    2(l + b) = 352

 ⇒   (l + b) = 352 = 176 2

Given ratio of sides = 9 : 7. So,

 Smallest side = 7 × 176 = 77 cm. 16

Hence, option A is correct.