NCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration InText Questions

NCERT Intext Question Page No. 170
Question 1.
(a) Match the following figures with their respective areas in the box.

(b) Write the perimeter of each shape.

(b)

(i)
The given figure is a rectangle in which
Length = 14cm
Breadth = 7cm
∵Perimeter of a rectangle = 2 × [Length + Breadth]
∴ Perimeter of the given figure = 2 × [14cm + 7cm]

(ii)

The figure is a square housing its side as 7cm. y
∵ Perimeter of a square = 4 × side
∴ Perimeter of the given figure = 4 × 7cm = 28cm

NCERT Intext Question Page No. 172
Question 1.
Nazma’s sister also has a trapezium shaped plot. Divide it into three parts as shown (Fig 11.4). Showthat the area of trapezium

Answer:
Area of ΔPWZ

Question 2.
If h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm, find the values of each of its parts separetely and add to find the area WXYZ. Verify it by putting the values of h, a and b in the expression h(a+b)2 .
Answer:

NCERT Intext Question Page No. 172
Question 1.
We know that parallelogram is also a quadrilateral. Let us also split such a quadrilateral into two triangles, find their areas and hence that of the parallelogram. Does this agree with the formula that you know already? (Fig 11.12)

Answer:
The diagonal BD of quadrilateral ABCD is joined and it divides the quadrilateral into two triangles.

Now,

NCERT Intext Question Page No. 175
Question 1.
A parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles?
Answer:
No, a trapezium cannot be divided into two congruent triangles.

NCERT Intext Question Page No. 176
Question 1.
Divide the following polygons (Fig 11.17) into parts (triangles and trapezium) to find out its area.

Answer:
(a) We draw perpendiculars from opposite vertices on FI, i.e. GL ⊥ FI, HM ⊥ FI and EN⊥ FI
Area of the polygon EFGHI
= ar (ΔGFL) + ar (trapezium GLMH) + ar (ΔHMI) + ar (ΔNEI) + ar (ΔEFN)

(b) NQ is a diagonal. Draw OA ⊥ NQ, MB ⊥ NQ, PC ⊥NQ and RD ⊥ NQ

∴ Area of Polygon OPQRMN
= ar (ΔOAN) + ar (trap. CPOA) + ar (ΔPCQ) + ar (ΔRDQ) + ar (trap. MBDR) + ar (ΔMBN)

Question 2.
Polygon ABCDE is divided into parts as shown below (Fig 11.18). Find its area if AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm.

Answer:
Area of polygon ABCDE = Area of ΔAFB + Area of ΔADE

So, the area of polygon ABCD = 3cm² + 7.5cm² + 3cm² + 10cm² = 23.5cm²