Subjective Questions
1. The area of a parallelogram is 392m2.f its altitude is twice the corresponding base, determine the
base and height.
2. The adjacent sides of a parallelogram are 36cm and 27cm in length If the distance between the
shorter sides is 12cm, find the distance between the longer sides.
3. A rectangular lawn, 75m by 60m, has two roads, each 4m wide, running through the middle of
the lawn, one parallel to length and other parallel, to breadth. Find the cost of gravelling the roads at
Rs 5.50 per m2
4. Using Heron's formula, find the area of an equilateral triangle if its side is' a 'units.
5. Find the percentage increase in the area of a triangle if its each side is doubled.
HOTS (High Order Thinking Skills) Question:
621 HEY DE BOLD
1. Find the area of quadrilateral ABCD whose sides in meters are 9,40,28 and 15 respectively and
the angle between first two sides is a right angle.
ilha Mittal
Class Teacher)
Answers
Answer:
Step-by-step explanation:
(1)Let the base(b) be 'x'
Given altitude is twice the base
It means,altitude(h) = 2x
Area of a parallelogram = bh
Given area of the parallelogram = 392 m²
⇒(bh)=392m²
⇒(x)(2x)=392m²
⇒2x²=392m²
⇒x²=392m²/2
⇒x²=196m²
⇒x=√196m²
⇒x=14m.
Therefore,the base of the parallelogram(x)=14m and
the altitude of the parallelogram(2x)=2(14)=28m
(2)measurement of the sides = 36 cm and 27 cm
perpendicular distance between shorter sides = 12cm
therefore, distance between longer sides = 36×h=27×12
= 36×h = 324
= h = 324/36
= 9cm .
(4)iven
that each sides of traingle is x cm
a=x
b=x
c=x
Semi-perrameter (s) =( a+b+ç)/2
=(x+x+x)/2
=3x/2
Area of traingle is = √{s(s-a)(s-b)(s-c)} (by Heron's formula.)
Area of traingle= √{s(s-x)(s-x)(s-x)}
Are of triangle= (√3/4)*x²
(5)Let a,b,c be the sides of the original ∆ & s be its semi perimeter.
S= (a+b+c)/2
2s= a+b+c.................(1)
The sides of a new ∆ are 2a,2b,2c
[ given: Side is doubled]
Let s' be the new semi perimeter.
s'= (2a+2b+2c)/2
s'= 2(a+b+c) /2
s'= a+b+c
S'= 2s. ( From eq 1)......(2)
Let ∆= area of original triangle
∆= √s(s-a)(s-b)(s-c).........(3)
&
∆'= area of new Triangle
∆' = √s'(s'-2a)(s'-2b)(s'-2c)
∆'= √ 2s(2s-2a)(2s-2b)(2s-2c)
[From eq. 2]
∆'= √ 2s×2(s-a)×2(s-b)×2(s-c)
= √16s(s-a)(s-b)(s-c)
∆'= 4 √s(s-a)(s-b)(s-c)
∆'= 4∆. (From eq (3))
Increase in the area of the triangle= ∆'- ∆= 4∆ - 1∆= 3∆
%increase in area= (increase in the area of the triangle/ original area of the triangle)× 100
% increase in area= (3∆/∆)×100
% increase in area= 3×100=300 %