by the Pythagoras theorem , h^2 = b^2+ A^2.AC = 15cm and AB = 20cm = BC= h.
by the Pythagoras theorem , h^2 = b^2+ A^2.AC = 15cm and AB = 20cm = BC= h.A/Q= h^2 = (20cm^2+15cm^2) = h^2 = 400cm + 225cm.
by the Pythagoras theorem , h^2 = b^2+ A^2.AC = 15cm and AB = 20cm = BC= h.A/Q= h^2 = (20cm^2+15cm^2) = h^2 = 400cm + 225cm.h^2= 625cm= h = √625cm = h = 25cm.
ANSWER IS :- BC = 25cm.
B.Step-by-step explanation:
It is given that,
It is given that,In ΔABC, AB = 15 cm, AC = 20 cm and ∠BAC = 90°
It is given that,In ΔABC, AB = 15 cm, AC = 20 cm and ∠BAC = 90°(i) Finding the area of ΔABC
It is given that,In ΔABC, AB = 15 cm, AC = 20 cm and ∠BAC = 90°(i) Finding the area of ΔABCIn Δ ABC, we have
It is given that,In ΔABC, AB = 15 cm, AC = 20 cm and ∠BAC = 90°(i) Finding the area of ΔABCIn Δ ABC, we haveBase, AB = 15 cm
Height, AC = 20 cm
Height, AC = 20 cm∴ Area of ΔABC = 1/2 * base * height = 1/2 * 15 * 20 = 150 cm²
(ii) Finding the length of AD
(ii) Finding the length of ADConsidering ΔABC and applying Pythagoras theorem, we get
(ii) Finding the length of ADConsidering ΔABC and applying Pythagoras theorem, we getBC = √[AC² + AB²] = √[20² + 15²] = √625 = 25 cm
(ii) Finding the length of ADConsidering ΔABC and applying Pythagoras theorem, we getBC = √[AC² + AB²] = √[20² + 15²] = √625 = 25 cmWe know that, if a perpendicular is drawn from the vertex containing a right angle of a right triangle to the hypotenuse then the triangle on each side of the perpendicular are similar to each other and to the whole triangle.
(ii) Finding the length of ADConsidering ΔABC and applying Pythagoras theorem, we getBC = √[AC² + AB²] = √[20² + 15²] = √625 = 25 cmWe know that, if a perpendicular is drawn from the vertex containing a right angle of a right triangle to the hypotenuse then the triangle on each side of the perpendicular are similar to each other and to the whole triangle.i.e., AD ⊥ BC
(ii) Finding the length of ADConsidering ΔABC and applying Pythagoras theorem, we getBC = √[AC² + AB²] = √[20² + 15²] = √625 = 25 cmWe know that, if a perpendicular is drawn from the vertex containing a right angle of a right triangle to the hypotenuse then the triangle on each side of the perpendicular are similar to each other and to the whole triangle.i.e., AD ⊥ BC∴ Δ ACD ~ ΔABD or ΔACD ~ Δ CBA or ΔABD ~ ΔCBA
(ii) Finding the length of ADConsidering ΔABC and applying Pythagoras theorem, we getBC = √[AC² + AB²] = √[20² + 15²] = √625 = 25 cmWe know that, if a perpendicular is drawn from the vertex containing a right angle of a right triangle to the hypotenuse then the triangle on each side of the perpendicular are similar to each other and to the whole triangle.i.e., AD ⊥ BC∴ Δ ACD ~ ΔABD or ΔACD ~ Δ CBA or ΔABD ~ ΔCBATaking ΔABD ~ ΔCBA, we have
..... [Corresponding sides of similar triangles are proportional to each other]
⇒ AD = [15 * 20] / 25
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABD
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABDConsidering ΔABD and applying Pythagoras theorem, we get
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABDConsidering ΔABD and applying Pythagoras theorem, we getBD = √[AB² - AD²] = √[15² - 12²] = √81 = 9 cm
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABDConsidering ΔABD and applying Pythagoras theorem, we getBD = √[AB² - AD²] = √[15² - 12²] = √81 = 9 cmIn ΔABD, we have
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABDConsidering ΔABD and applying Pythagoras theorem, we getBD = √[AB² - AD²] = √[15² - 12²] = √81 = 9 cmIn ΔABD, we haveBase, BD = 9 cm
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABDConsidering ΔABD and applying Pythagoras theorem, we getBD = √[AB² - AD²] = √[15² - 12²] = √81 = 9 cmIn ΔABD, we haveBase, BD = 9 cmHeight, AD = 12 cm
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABDConsidering ΔABD and applying Pythagoras theorem, we getBD = √[AB² - AD²] = √[15² - 12²] = √81 = 9 cmIn ΔABD, we haveBase, BD = 9 cmHeight, AD = 12 cm∴ Area of ΔABD = 1/2 * base * height = 1/2 * 12 * 9 = 54 cm²
Jawaban:
A.Answer:
by the Pythagoras theorem , h^2 = b^2+ A^2.
by the Pythagoras theorem , h^2 = b^2+ A^2.AC = 15cm and AB = 20cm = BC= h.
by the Pythagoras theorem , h^2 = b^2+ A^2.AC = 15cm and AB = 20cm = BC= h.A/Q= h^2 = (20cm^2+15cm^2) = h^2 = 400cm + 225cm.
by the Pythagoras theorem , h^2 = b^2+ A^2.AC = 15cm and AB = 20cm = BC= h.A/Q= h^2 = (20cm^2+15cm^2) = h^2 = 400cm + 225cm.h^2= 625cm= h = √625cm = h = 25cm.
ANSWER IS :- BC = 25cm.
B.Step-by-step explanation:
It is given that,
It is given that,In ΔABC, AB = 15 cm, AC = 20 cm and ∠BAC = 90°
It is given that,In ΔABC, AB = 15 cm, AC = 20 cm and ∠BAC = 90°(i) Finding the area of ΔABC
It is given that,In ΔABC, AB = 15 cm, AC = 20 cm and ∠BAC = 90°(i) Finding the area of ΔABCIn Δ ABC, we have
It is given that,In ΔABC, AB = 15 cm, AC = 20 cm and ∠BAC = 90°(i) Finding the area of ΔABCIn Δ ABC, we haveBase, AB = 15 cm
Height, AC = 20 cm
Height, AC = 20 cm∴ Area of ΔABC = 1/2 * base * height = 1/2 * 15 * 20 = 150 cm²
(ii) Finding the length of AD
(ii) Finding the length of ADConsidering ΔABC and applying Pythagoras theorem, we get
(ii) Finding the length of ADConsidering ΔABC and applying Pythagoras theorem, we getBC = √[AC² + AB²] = √[20² + 15²] = √625 = 25 cm
(ii) Finding the length of ADConsidering ΔABC and applying Pythagoras theorem, we getBC = √[AC² + AB²] = √[20² + 15²] = √625 = 25 cmWe know that, if a perpendicular is drawn from the vertex containing a right angle of a right triangle to the hypotenuse then the triangle on each side of the perpendicular are similar to each other and to the whole triangle.
(ii) Finding the length of ADConsidering ΔABC and applying Pythagoras theorem, we getBC = √[AC² + AB²] = √[20² + 15²] = √625 = 25 cmWe know that, if a perpendicular is drawn from the vertex containing a right angle of a right triangle to the hypotenuse then the triangle on each side of the perpendicular are similar to each other and to the whole triangle.i.e., AD ⊥ BC
(ii) Finding the length of ADConsidering ΔABC and applying Pythagoras theorem, we getBC = √[AC² + AB²] = √[20² + 15²] = √625 = 25 cmWe know that, if a perpendicular is drawn from the vertex containing a right angle of a right triangle to the hypotenuse then the triangle on each side of the perpendicular are similar to each other and to the whole triangle.i.e., AD ⊥ BC∴ Δ ACD ~ ΔABD or ΔACD ~ Δ CBA or ΔABD ~ ΔCBA
(ii) Finding the length of ADConsidering ΔABC and applying Pythagoras theorem, we getBC = √[AC² + AB²] = √[20² + 15²] = √625 = 25 cmWe know that, if a perpendicular is drawn from the vertex containing a right angle of a right triangle to the hypotenuse then the triangle on each side of the perpendicular are similar to each other and to the whole triangle.i.e., AD ⊥ BC∴ Δ ACD ~ ΔABD or ΔACD ~ Δ CBA or ΔABD ~ ΔCBATaking ΔABD ~ ΔCBA, we have
..... [Corresponding sides of similar triangles are proportional to each other]
⇒ AD = [15 * 20] / 25
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABD
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABDConsidering ΔABD and applying Pythagoras theorem, we get
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABDConsidering ΔABD and applying Pythagoras theorem, we getBD = √[AB² - AD²] = √[15² - 12²] = √81 = 9 cm
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABDConsidering ΔABD and applying Pythagoras theorem, we getBD = √[AB² - AD²] = √[15² - 12²] = √81 = 9 cmIn ΔABD, we have
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABDConsidering ΔABD and applying Pythagoras theorem, we getBD = √[AB² - AD²] = √[15² - 12²] = √81 = 9 cmIn ΔABD, we haveBase, BD = 9 cm
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABDConsidering ΔABD and applying Pythagoras theorem, we getBD = √[AB² - AD²] = √[15² - 12²] = √81 = 9 cmIn ΔABD, we haveBase, BD = 9 cmHeight, AD = 12 cm
⇒ AD = [15 * 20] / 25⇒ AD = 12 cm(iii) Area of ΔABDConsidering ΔABD and applying Pythagoras theorem, we getBD = √[AB² - AD²] = √[15² - 12²] = √81 = 9 cmIn ΔABD, we haveBase, BD = 9 cmHeight, AD = 12 cm∴ Area of ΔABD = 1/2 * base * height = 1/2 * 12 * 9 = 54 cm²
Sorry can only answer a and b:)
Hope it helps sorry if it's wrong:)