Math, asked by singhgurveer0747, 5 months ago

The equal sides of the isosceles triangle are 12 cm, and the perimeter is 30 cm. The area of this triangle is: ਜੇਕਰ ਸਮਦੋਭੁਜੀ ਤ੍ਰਿਭੁਜ ਦੀਆਂ ਬਰਾਬਰ ਭੁਜਾਵਾਂ ਵਿਚੋਂ ਹਰੇਕ ਦੀ ਲੰਬਾਈ 12 cm ਹੋਵੇ ਅਤੇ ਪਰਿਮਾਪ 30 cm ਹੋਵੇ I ਤਾਂ ਤ੍ਰਿਭੁਜ ਦਾ ਖੇਤਰਫਲ ਪਤਾ ਕਰੋ Iयदि एक समद्विबाहु त्रिभुज की बराबर भुजाएँ में से हरेक 12 सेमी हैं और परिमाप 30 सेमी है, तो त्रिभुज का क्षेत्रफल ज्ञात कीजिए। *

9√15 ਸਮ² (cm²)

6√15 ਸਮ² (cm²)

3√15 ਸਮ² (cm²)

√15 ਸਮ² (cm²)

Answers

Answered by udayteja5660
4

Answer:

9√15 cm²

Step-by-step explanation:

Let the given isosceles triangle be ΔABC

Given

The perimeter of the triangle = 30 cm

We know that

The perimeter of the triangle =  side₁ + side₂ + side₃ = AB + BC + AC

In the case of an isosceles triangle, two sides are equal.(AB = AC)

The length of the equal sides of the isosceles triangle = a = 12 cm.

⇒AB = AC = 12 cm

Let the base of the triangle be BC = 'x' cm.

⇒ a + a + x = 30 cm

⇒2a + x = 30 cm

⇒2(12) + x = 30 cm

⇒x + 24 = 30

⇒x = 30 - 24

⇒x = 6 cm

So the length of the base of the trianlge BC = 6 cm.

We draw a perpendicular from A to the base BC at D.

D is mid-point of BC.

⇒BD = DC = BC/2

⇒BD = DC = 6/2

BD = DC = 3 cm

So AD is the height of the ΔABC and AD⊥BC.

AD divides ΔABC into two right-angled triangles ΔADB and ΔADC.

Let us take ΔADC.

Here AD and DC are the sides and AC is the hypotenuse of the right-angled ΔADC.

By Pythagoras theorem,

Side² + side² = hypotenuse²

⇒AD² + DC² = AC²

⇒AD² + 3² = 12²         [∵CD = 3 cm, AC = 12 cm]

⇒AD² + 9 = 144

⇒AD² = 144 - 9

⇒AD² = 135

⇒AD² = 15*9

⇒AD = √(15*9)

∴AD = 3√15 cm

Height of the ΔABC = 3√15 cm

Base of the ΔABC = 6 cm

∴Area of the ΔABC = 1/2*(base)*(height) sq.units

                                = 1/2*(6)*(3√15) cm²

                                = 3*3√15 cm²

∴Area of the ΔABC = 9√15 cm²

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