Question

PQRS is a parallelogram of area 35 cm^2. T is a point on QR such that QT = 4 cm and TR = 3 cm. Calculate the area of ΔPQT.

Answer 1

Step-by-step explanation:

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Answer 2

$\Large {\underline { \sf {Explication \: of \: Steps :}}}$

Given that,

• PQRS is a parallelogram.
• Area of parallelogram is 35 cm².
• T is a point on QR such that QT = 4 cm and TR = 3 cm.

We have to find the area of ∆PQT.

Let PO be the height of the triangle and the parallelogram.

• Height of ∆PQT = Height of ||gm PQRS

Finding height of the parallelogram PQRS :

We know that,

★ Area of parallelogram = Base × Height

• Base = QR

⇒ QR = QT + TR

⇒ QR = 4 cm + 3 cm

⇒ QR = 7 cm

⇒Base = 7 cm

Substituting value of base and the area of parallelogram to find the height.

$\longmapsto$ 35 cm² = 7 cm × (h) cm

$\longmapsto$ $\sf \dfrac{35\: cm^2 }{7\: cm}$ = (h) cm

$\longmapsto$ 5 cm = (h) cm

$\longmapsto \sf \red {Height = 5 \: cm }$

So, the height of the parallelogram (PO) is 5 cm. PO is also the height of ∆PQT. So, height of the triangle ∆PQT is 5 cm.

In ∆ PQT :

• Base = QT ⇒ 4 cm
• Height = PO ⇒ 5 cm

We know that,

★ Area of ∆ = $\sf \dfrac{1}{2}$ × b × h

$\longmapsto$ Area of ∆PQT = $\sf \dfrac{1}{2}$ × 4 cm × 5 cm

$\longmapsto$ Area of ∆PQT = 1 × 2 cm × 5 cm

$\longmapsto \sf \red {Area \: of \: \Delta PQT = 10 \: cm^2 }$

Therefore, area of the ∆PQT is 10 cm².