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
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