Question

triangle ABC has side AB =7.5 , AC=6.5cm and BC=7cm on the base of BC. a parallelogram is DBCE of same area as that of triangle ABC is constructed find the height DF of the parallelogram ​

Answer 1

Answer:

3 cm

Step-by-step explanation:

h = 3 cm

Answer 2

$\underline\mathfrak\pink{Given:-}$

• ∆ ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm.

$\large\underline\mathfrak{To \: find:-}$

• Find the height of the parallelogram ....?

$\large\underline\mathfrak\pink{Solutions:-}$

Area of ∆ABC

The sides of a triangle are:-

• AB = 7.5 cm
• AC = 6.5 cm
• BC = 7 cm

$\: \: \: \: \: \therefore \: \: Area \: \: of \: \: triangle \: \: by \: \: heroes's \: \: formula:-$

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

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{a \: + \: b \: + \: c}{2}$

• a ⇢ 7.5 cm

• b ⇢ 6.5 cm

• c ⇢ 7 cm

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{{7.5} \: + \: {7} \: + \: {6.5}}{2}$

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{21}{2}$

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: {10.5}$

$\: \: \: \: \: \therefore \: \: \: Area \: \: of \: \: triangular \: \: field:-$

$\: \: \: \: \: \leadsto \: \: \sqrt{{10.5} \: ({10.5} \: - \: {7.5}) \: ({10.5} \: - \: {7}) \: ({10.5} \: - \: {6.5})}$

$\: \: \: \: \: \leadsto \: \: \sqrt{{10.5} \: \times \: {(3)} \: \times \: {(3.5)} \: \times \: {(4)}}$

$\: \: \: \: \: \leadsto \: \: \sqrt{441}$

$\: \: \: \: \: \leadsto \: \: {21} \: {cm}^{2}$

$\: \: \: \: \: \: \: \: Area \: \: of \: \: triangle \: \: \leadsto \: \: {96.82} \: {cm}^{2}$

Now,

$\: \: \: \: \: \: \: \: Area \: \: of \: \: parallelogram \: \: BCED \: \: = \: \: Base \: \times \: height$

$\: \: \: \: \: \: \: \: \leadsto \: \: BC \: \: \times \: \: DF$

$\: \: \: \: \: \: \: \: \leadsto \: \: {7} \: \: \times \: \: DF$

$\: \: \: \: \: \: \: \: Area \: \: of \: \: \triangle \: ABC \: \: = \: \: Area \: \: of \: \: parallelogram \: \: BCED$

$\: \: \: \: \: \: \: \: \leadsto \: \: {21} \: \: = \: \: {7} \: \: \times \: \: DF$

$\: \: \: \: \: \: \: \: \leadsto \: \: {DF} \: \: = \: \: \frac{21}{7}$

$\: \: \: \: \: \: \: \: \leadsto \: \: {DF} \: \: = \: \: {3} \: \: cm$

$\: \: \: \: \: \: \: \: Hence, \\ \: \: The \: \: height \: \: of \: \: parallelogram \: \: is \: \: {3} \: {cm}$

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