Question

Q4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.

Answer 1

Step-by-step explanation:

Area of a triangle =

$\sqrt{s(s - a)(s - b)(s - c)}$

• So let The sides be
• a=13
• b=14
• c=15

S is the semi-perimetre

Therefore,

$s = \frac{(a + b + c)}{2}$

$s = \frac{13 + 14 +15 }{2} = \frac{42}{2} = 21 \: sqcm$

Now

$\ = sqrt{21( 21 - 13)(21 - 14)(21 - 15)} \\ \ = sqrt{21(8)(7)(6)} \\ = \sqrt{7056} \\ = 84 \: sqcm$

SO,

Area of triangle = 84sqcm. ......EQ 1

Area of Parallelogram

$= \frac{1}{2} (b \times h)$

Where,

• b= Base= 14 cm
• h = Hieght

So,

$= \frac{1}{2} (14 \times h) \\ =7h$

So,

Area of Parallelogram = 7h. .....EQ 2

But we know

$area \: of \: triangle = \: \\ </p><p> area \: of \: paralleogram$

Therefore, From EQ 1 and 2

$7h = 84 \\ h = \frac{84}{7} \\ h = 12cm$

Therefore, Hieght of the Parallelogram (H) = 12cm

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