RD Sharma Test: Triangles - Grade 10 MCQ

Let two isosceles triangles be ΔABC and ΔDEF And their corresponding altitudes are AM and DN.
Since the corresponding angles of both triangles are equal.

Since, in a triangle the sum of squares of any wo sides is equal to twice the square of half of the third side together with twice the square of the median bisecting it.
In ΔPQM,

If in two triangles ABC and DEF, 

ΔFDE∼ΔCAB because for similarity, all sides should be in proportion.

Since the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

AB = 13 m Therefore, the distance between their tops is 13 m. 

If ΔABC ∼ ΔEDF and ΔABC is not similar to ΔDEF,ΔDEF,then BC. DE = AB .EF 
Here according to given condition, BC. DE  ≠ AB .EF

Given: AC || BD.and AC = 6 cm, AE = 3 cm, EB = 4 cm, ED = 8 cm,
In triangles ACE and DEB, ∠AEC = ∠DEB [Vertically opposite angles] ∠ECA = ∠EDB [Alternate angles as

The diameter of circle is AC. Here ∠ABC is angle of semicircle. 

AB ∴ ∠ ABC = 90° ∴ Δ ABC is aright angled triangle. 

10 cm

Let ABCD be a rhombus.
One diagonal BC = 8 cm and dise BC 5 cm
Since diagonals of a rhombus bisect each other at 90^∘

Since, Sides of ΔABC and ΔPQR are proportional, therefore, .
∴ ∠A = ∠Q, ∠B = ∠R and ∠C = ∠P
⇒ ∠P
= 20° and ∠R
​= 30° 

In triangles APB and CPD,
∠APB = ∠CPD [Vertically opposite angles] ∠BAP = ∠ACD [Alternate angles as AB||CD]

In ΔPOQ and ΔCOB, ∠PQO = ∠QBO
[Alternate angles] ∠POO = ∠COB [Vertically Opposite] 
.[AA similarity] 

Let ABCD be the required rhombus whose diagonals are AC = 10 cm and BD = 24 cm. Let AC and BD intersect each other at O.
Then AB = BC = CD = DA
Also AO = CO = 10/2 = 5 cm
And BO = DO = 24/2 = 12 cm

Let the two similar triangles be ΔABC and ΔDEF such that ar(ΔABC) = 100 cm² and ar(ΔDEF) = 49 cm².
Let AM and DN be the respective altitudes of ΔABC and ΔDEF.
Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding altitudes

Using Pythagoras Theorem,
AC² = AB² + BC² = 64 + 225 = 289 8 m ⇒ AC 17 cm