Question

in the diagram PQRST is a rhombus and TSR is a straight line calculate the area of the whole diagram.

help me with this please I have exams tmrw


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Answer 1

✬ Area = 104 cm² ✬

Step-by-step explanation:

Given:

• PQST is a rhombus.
• TSR is a straight line.
• QRS is a right angle triangle at R.

To Find:

• What's is the area of whole figure ?

Solution: Let the length of SR be x cm.

As we know that

• All sides of a rhombus are equal and opposite sides are parallel.

∴ PQ = QS = ST = TP = 10 cm

∴ Height of PQST = QR = 8 cm

• So each side of rhombus will be 10 cm and height will be 8 cm

★ Area of Rhombus = Base × Height ★

➟ Ar. PQST = TS × QR

➟ (10 × 8) cm²

➟ 80 cm²

Now in ∆QRS , we have

• QR = 8 cm {perpendicular}

• SR = Base

• QS = 10 cm {hypotenuse}

Applying Pythagoras Theorem

★ Hypotenuse² = P² + B² ★

➨ 10² = 8² + x²

➨ 100 – 64 = x²

➨ √36 = x

➨ 6 = x

So length of base of ∆ is 6 cm.

★ Area of ∆ = 1/2 × Base × Height ★

➼ Area ∆QSR = 1/2 × 6 × 8

➼ (3 × 8) cm²

➼ 24 cm²

∴ Total area of figure = Area (∆ + Rhombus)

$\implies{\rm }$ 24 + 80

$\implies{\rm }$ 104 cm²

Hence, the area of whole diagram is 104 cm².

Answer 2

Answer:

Given :-

• PQRST is a rhombus are TSR is a straight line.

To Find :-

• What is the area of the whole diagram or figure.

Solution :-

First, we have to find the side of rhombus :

Given :

• PQRST is a rhombus.

Then,

$\implies \sf Side\: of\: Rhombus =\: PQ = QS = ST = ST = TP$

$\implies \sf\bold{\green{Side\: of\: Rhombus =\: 10\: cm}}$

Hence, the side of rhombus is 10 cm .

Now, we have to find the height of rhombus :

$\implies \sf Height\: of\: Rhombus =\: QR$

$\implies\sf\bold{\green{Height\: of\: Rhombus =\: 8\: cm}}$

Hence, the height of rhombus is 8 cm .

Now, we have to find the area of rhombus :

As we know that,

$\clubsuit$ Area of Rhombus :

$\longmapsto \sf\boxed{\bold{\pink{Area\: of\: Rhombus =\: Base \times Height}}}$

Given :

• Base (TQ) = 10 cm
• Height (QR) = 8 cm

According to the question by using the formula we get,

$\implies \sf Area\: of\: Rhombus =\: TQ \times QR$

$\implies \sf Area\: of\: Rhombus =\: 10\: cm \times 8\: cm$

$\implies \sf\bold{\purple{Area\: of\: Rhombus =\: 80\: {cm}^{2}}}$

Now, we have to find the length of the base :

Let,

$\mapsto$ Length of the base = x cm

As we know that,

$\clubsuit$ Pythagoras Theorem :

$\longmapsto \sf\boxed{\bold{\pink{{(Hypotenuse)}^{2} =\: {(Perpendicular)}^{2} + {(Base)}^{2}}}}$

Given :

• Hypotenuse (QS) = 10 cm
• Perpendicular (QR) = 8 cm

According to the question by using the formula we get,

$\implies \sf {(10)}^{2} =\: {(8)}^{2} + {(x)}^{2}$

$\implies \sf 100 =\: 64 + {(x)}^{2}$

$\implies \sf - {(x)}^{2} =\: 64 - 100$

$\implies \sf {\cancel{-}} {(x)}^{2} =\: {\cancel{-}} 36$

$\implies \sf {(x)}^{2} =\: 36$

$\implies \sf x =\: \sqrt{36}$

$\implies \sf\bold{\purple{x =\: 6\: cm}}$

Hence, the length of the base is 6 cm .

Now, we have to find the area of an triangle :

As we know that,

$\clubsuit$ Area of Triangle :

$\longmapsto \sf\boxed{\bold{\pink{Area\: of\: triangle =\: \dfrac{1}{2} \times Height \times Base}}}$

Given :

• Height (QR) = 8 cm
• Base (SR) = 6 cm

According to the question by using the formula we get,

$\implies \sf Area\: of\: triangle =\: \dfrac{1}{2} \times 8 \times 6$

$\implies\sf Area\: of\: triangle =\: \dfrac{1}{\cancel{2}} \times {\cancel{48}}$

$\implies \sf Area\: of\: triangle =\: 1 \times 24$

$\implies \sf\bold{\purple{Area\: of\: traingle =\: 24\: {cm}^{2}}}$

Atlast, we have to find the area of whole figure :

$\implies \sf \bold{Area\: of\: whole\: figure =\: Area\: of\: triangle + Area\: of\: Rhombus}$

Given :

• Area of Triangle = 24 cm²
• Area of Rhombus = 80 cm²

$\implies \sf Area\: of\: whole\: figure =\: 24\: {cm}^{2} + 80\: {cm}^{2}$

$\implies\sf\bold{\red{Area\: of\: whole\: figure =\: 104\: {cm}^{2}}}$

$\therefore$ The area of the whole figure is 104 cm².