S.chand publication New Learning Composite mathematics solution of class 7 Chapter 14 Perimeter and Area Exercise 14D

 Exercise 14D

Question 1

Complete the table for a triangle. All measurement are in centimetres 

	(a)	(b)	(c)	(d)	(e)	(f)	(g)
Base	6	10	7	8	12	13	18
Height	4	68	4	10	9	9	15
Area	12	34	14	40	54	58.5	135

Question 2

Find in square metres the area of a triangle whose

(a) base=8 cm, altitude=6 cm

Area of the triangle$=\left(\frac{1}{2}\times 8 \times 6\right)$ sq cm

=24 sq cm

=0.0024 sq m

(b) base=38000 mm, altitude=4500 mm

Area of the triangle$=\frac{1}{2} \times 38000 \times 4500$ sq mm

=85500000

=85.5 sq m

Question 3

Find the area of shaded portion of each of the following figures (measure in cm)

(a) Figure to be added

Area of ▭EBCD=(18×25)cm²

=450

In ΔAED, AE=(30-25)cm

=5

ED=18 cm

∴Area of ΔAED$=\left(\frac{1}{2} \times 5 \times 18\right)$cm²

=45 cm²

∴Area of the shaded figure=(450+45) cm²

=495 cm²

(b) Figure to be added

∴Area of ΔABC$=\left(\frac{1}{2} \times 6 \times 15\right)$sq cm

=45

∴Area of ▭ACDF=(15×11)sq cm

=165

∴Area of ΔACDF$=\left(\frac{1}{6} \times 6 \times 15\right)$ sq cm

=45

∴Area of the shaded figure=(45+165+45)sq m

=255

(c) Figure to be added

In ▭AFGE, AE=FG=11 cm

AF=EG=9 cm

∴Area of ▭AFGE=(11×9)cm²

=99 cm²

In ΔFBC, $FC=\frac{11}{2}$ cm

FB=(15-9)cm

=6 cm

∴Area of ΔFBC$=\left(\frac{1}{2} \times \frac{11}{2} \times 6\right)$ cm²

=16.5 cm²

In ΔCDG, $CG=\frac{11}{2}$ cm

GD=(15-9)=6 cm

∴Area of ΔCDG, $CG=\frac{11}{2}$ cm

GD=(15-9) cm

=6 cm

∴Area of ΔCDG$=\left(\frac{1}{2} \times \frac{11}{2} \times 6\right)$ cm²

=16.5

∴Area of the shaded figure=(99+16.5+16.5) cm²

=132

(d) Figure to be added

∴Area of  ΔABC$=\left(\frac{1}{2} \times 10 \times 15\right)$ cm²

=75

∴Area of ΔCDE$=\left(\frac{1}{2} \times 10 \times 15\right)$ cm²

=75 cm²

(e) Figure to be added

In ΔABC, AC=(6+6+6) cm

=18 cm

altitude= 10 cm

∴Area of ΔABC$=\left(\frac{1}{2} \times 18 \times 10\right)$ cm²

=90 cm²

Area of ▭DEFG=(28×6) cm²

=168

∴Area of shaded figure=(90+168) cm²

=258

(f) Figure to be added

In ΔABF, AB=(16-2.5) cm

=13.5 cm

BF=(13-5)=8 cm

∴Area of ΔABF$=\left(\frac{1}{2} \times 8 \times 13.5\right)$ cm²

=54

∴Area of shaded figure=(54+32.5) cm²

=86.5 cm²

Question 4

The area of  a triangle is equal ti that of a square of side length 40 cm. Find the side of the triangle whose corresponding altitude is 64 cm

Side of the square=40 cm

Area of the square=(40)² cm²

Let, side of the triangle be b

ATQ,

$\frac{1}{2} \times b \times 64=1600$

$b=\frac{1600 \times 2}{64}$

∴b=50

Ans : Side of the triangle is 50 cm

Question 5

A rectangular field is 28 m long and 18 m wide. How many triangular flower beds each base 7m and altitude 6m can be laid in the field?

∴Area of the rectangular field=(28×18) m²

=504 m²

∴Area of the triangle$=\left(\frac{1}{2} \times 7 \times 6\right)$ m²

=21 m²

∴Number of triangular flowerbeds can be laid in the field$=\frac{504}{21}=24$

Ans : 24 triangular flowerbeds can be laid in the field