Question

1. The base of an isosceles triangle is 16 cm and its area is 48 cm^2. The perimeter of the triangle is

(a) 41 cm
(b) 36 cm
(c) 48 cm
(d) 324 cm

2. The area of an equilateral triangle is 36√3 cm'. Its perimeter is

(a) 36 cm
(b) 12√3 cm
(c) 24 cm
(d) 30 cm

3. Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is

(a) 156 cm
(b) 78 cm
(c) 60 cm
(d) 120 cm

4. The base of a right triangle is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is

(a) 168 cm
(b) 252 cm
(c) 336 cm
(d) 504 cm

5. The area of an equilateral triangle is 81/3 cm . Its height is

(a) 9√3 cm
(b) 6√3 cm
(c) 18√3 cm
(d) 9 cm

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Answer 1

Answer:

1. a.

2just

Step-by-step explanation:

1a,2b,3c,4d,5d

Answer 2

1. The base of an isosceles triangle is 16 cm and its area is 48 cm^2. The perimeter of the triangle is

(a) 41 cm

(b) 36 cm

(c) 48 cm

(d) 324 cm

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$\bf \:\large \red{AηsωeR : 1.} ✍$

Given :-

☆In triangle ABC,

☆Base of isosceles triangle, BC = 16 cm

☆Area of triangle ABC = 48 sq. cm

☆Let height of triangle, AD be 'h' cm.

☆Using Formula,

$\boxed{\bf \:Area \: of \: triangle = \dfrac{1}{2} \times base \times height}$

$\bf\implies \:48 = \dfrac{1}{2} \times 16 \times h$

$\bf\implies \:48 = 8h$

$\bf\implies \:h = 6 \: cm$

☆Since, perpendicular AD divides the base BC in two equal parts.

☆So, BD = DC = 8cm

☆Now, In right triangle ABD.

☆Using Pythagoras Theorem,

$\begin{gathered}{\boxed{\sf{\pink{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}}}\end{gathered}$

$\bf\implies \: {AB}^{2} = {BD}^{2} + {AD}^{2}$

$\bf \:  ⟼ {AB}^{2} = {8}^{2} + {6}^{2}$

$\bf \:  ⟼ {AB}^{2} = 36 + 64$

$\bf \:  ⟼ {AB}^{2} = 100$

$\bf\implies \:AB = 10 \: cm$

☆Since, triangle ABC is an isosceles triangle.

☆Therefore, AB = AC.

☆So, Perimeter of Triangle ABC = AB + BC + CA

☆So, Perimeter of Triangle ABC = 10 + 10 + 16 = 36 cm

$\large{\boxed{\boxed{\bf{Option \: (b) \: is \: correct}}}}$

$\bf \:↝ For \: figure \: see \: the \:  first \: attachment.$

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2. The area of an equilateral triangle is 36√3 cm'. Its perimeter is

(a) 36 cm

(b) 12√3 cm

(c) 24 cm

(d) 30 cm

$\bf \:\large \red{AηsωeR : 2.} ✍$

☆Area of equilateral triangle = 36√3 sq. cm

☆Let side of triangle be 'x' cm

☆So, using formula of area of triangle,

$\boxed{\bf \:Area \: of \: triangle = \dfrac{ \sqrt{3} }{4} {x}^{2} }$

$\bf\implies \:36 \sqrt{3} = \dfrac{ \sqrt{3} }{4} \times {x}^{2}$

$\bf\implies \: {x}^{2} = 36 \times 4$

$\bf\implies \:x = 6 \times 2 = 12 \: cm$

☆So, Perimeter of Triangle = 3x = 3 × 12 = 36 cm

$\large{\boxed{\boxed{\bf{Option \: (a) \: is \: correct}}}}$

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3. Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is

(a) 156 cm

(b) 78 cm

(c) 60 cm

(d) 120 cm

$\bf \:\large \red{AηsωeR : 3.} ✍$

☆Let us consider a triangle ABC in which AB = AC = 13 cm and BC = 24 cm.

☆Let AD is perpendicular to BC.

☆Since, AD divides the base BC in to two equal parts.

☆Therefore, BD = DC = 12 cm.

☆In right angle triangle ABD,

☆Using Pythagoras Theorem,

$\begin{gathered}{\boxed{\sf{\pink{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}}}\end{gathered}$

$\bf\implies \: {AB}^{2} = {BD}^{2} + {AD}^{2}$

$\bf\implies \: {13}^{2} = {AD}^{2} + {12}^{2}$

$\bf \:  ⟼ 169 = {AD}^{2} + 144$

$\bf \:  ⟼ {AD}^{2} = 169 - 144$

$\bf \:  ⟼ {AD}^{2} = 25$

$\bf \:  ⟼ AD = 5 \: cm$

☆Now, area of triangle ABC is given by

$\boxed{\bf \:Area \: of \: triangle = \dfrac{1}{2} \times base \times height}$

${\bf \:Area \: of \: triangle = \dfrac{1}{2} \times 24 \times 5 = 60 \: {cm}^{2} }$

$\large{\boxed{\boxed{\bf{Option \: (c) \: is \: correct}}}}$

$\bf \:↝ For \: figure \: see \: the \:   {2}^{nd} \: attachment.$

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4. The base of a right triangle is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is

(a) 168 cm

(b) 252 cm

(c) 336 cm

(d) 504 cm

$\bf \:\large \red{AηsωeR : 4.} ✍$

☆Let us consider a right triangle ABC, right angled at B, so that

☆Base, BC = 48 cm

☆Hypotenuse, AC = 50 cm

☆Using Pythagoras Theorem,

$\begin{gathered}{\boxed{\sf{\pink{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}}}\end{gathered}$

$\bf \:  ⟼ {AC}^{2} = {AB}^{2} + {BC}^{2}$

$\bf \:  ⟼ {50}^{2} = {48}^{2} + {AB}^{2}$

$\bf \:  ⟼ {AB}^{2} = 2500 - 2304$

$\bf \:  ⟼ {AB}^{2} = 196$

$\bf \:  ⟼ AB = 14 \: cm$

$\boxed{\bf \:Area \: of \: triangle = \dfrac{1}{2} \times base \times height}$

$\bf \:  ⟼ Area \: of \: triangle =\dfrac{1}{2} \times AB \times BC$

$\bf \:  ⟼ Area \: of \: triangle = \dfrac{1}{2} \times 14 \times 48$

$\bf \:  ⟼ Area \: of \: triangle =336 \: {cm}^{2}$

$\large{\boxed{\boxed{\bf{Option \: (c) \: is \: correct}}}}$

$\bf \:↝ For \: figure \: see \: the \: {3}^{rd} \: attachment.$

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