11. An isosceles triangle has perimeter 30 cm and each of the equal is 12 cm. Find the area o
qual is 12 cm. Find the area of the triangle.
12. Find the area of an equilateral triangle whose each side is of 'a' units. Also, find its altitude
13. Find the area of the quadrilateral field ABCD in which AB = 35 m, BC = 75 m, CD = 60 m
in which ADsm, BCD m, CD = 60 m. DA= 66 m and
ZCBD = 90°
14. The adjacent sides of a parallelogram ABCD are AB = 34 cm, BC = 20 cm and its diagonal AC = 42 cm.
Find the area of the parallelogram.
15. The sides of a triangle are 39 cm, 42 cm and 45 cm. A parallelogram stands on the greatest side of the
triangle and has the same area as that of the triangle. Find the height of the parallelogram.
Answers
Answer:
for all the above question use herons formula
Step-by-step explanation:
11. Let the third side be x.
Two equal sides measure is 12 cm
•.• Perimeter of ∆
= {(2 × equal sides) + (third side)} cm = {(2 × 12) + x} cm
= (24 + x) cm
But, the given perimeter = 30 cm.
A/q,
24 + x = 30
x = 30 - 24
x = 6
Hence, Length of third side = 6 cm
Now, height of ∆
Let ∆ABD the right angled ∆ with AB as the hypotenus.
Then, (AB)^2 = (AD)^2 + (BD)^2 [Py.the.] (12)^2 = (AD)^2 + (3)^2
144 = (AD)^2 + 9
144 - 9 = (AD)^2
( AD)^2 = 135
AD = √3 × 3 × 3 × 5
AD = 3√15 cm
Hence, height of ∆ = 3√15 cm
✓✓Area of the ∆
= (1/2 × base × height) sq. unit
= (1/2 × 6 × 3√15 ) cm^2
= 9√15 cm^2
= 34.85 cm^2 ~ 35 cm^2
12. Side of equilateral ∆ = a unit
Hence, area of the ∆ = (√3)a^2/4 = 1.732 a^2 /4 =0.433 a^2 unit^2
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