RS Aggarwal Test: Triangles - Grade 10 MCQ

The angle bisector theoremstates that an angle bisector divides the opposite side of a triangle into two segments that are proportional to the triangle's other two sides. In other words,
AB/BD = AC/CD.
Now
AB/AC=BD/DC
Which is the required ratio .
Thats how BD/DC=3/4

Δ ABC is an equilateral triangle

By Pythagoras theorem in triangle ABD

AB² = AD² + BD²

but BD = 1/2 BC (∵ In a triangle, the perpendicular from the vertex to the base bisects the base)

thus AB² = AD² + {1/2 BC}²

AB² = AD² + 1/4 BC²

4 AB² = 4AD² + BC²

4 AB² - BC² = 4 AD²

(as AB = BC we can subtract them )

Thus 3AB² = 4AD²

Given :△ABC, D, E and F are mid points of AB, BC, CA respectively.
Using mid point theorem we prove that □ADEF, □DBEF and □DECF are parallelograms. The diagonal of a parallelogram divides the parallelogram into two congruent triangles. So all triangles are congruent to each other. And each small triangle is similar to the original triangle.

Angles of ΔABC:

• ∠A = 50°
• ∠B = 70°
• ∠C = 60°

Angles of ΔDEF:

• ∠D = 60°
• ∠E = 70°
• ∠F = 50°

We can observe the following:

• ∠A = 50° corresponds to ∠F = 50°
• ∠B = 70° corresponds to ∠E = 70°
• ∠C = 60° corresponds to ∠D = 60°

So, the corresponding angles of ΔABC and ΔDEF are equal, meaning the two triangles are similar.

The correct answer is:

d) ΔFED.

The correct option is Option D.

Given : D, E and F are tge mid-point of the sides BC, CA and AB respectively of Δ ABC.

To prove : Δ DEF is congruent to traingle

Proof : 

Since E and F are midpoints of AC and AB. 

BC II FE and FE = ½ BC = BD (By mid point theorem)

BD II FE and BD = FE 

Similarly, BF II DE and BF = DE

Hence, BDEF is a parallelogram (A pair of opposite sides are equal and parallel)

Similarly, we can prove that FDCE and AFDE are also parallelograms. 

Now, BDEF is a parallelogram so it's diagonal FD divides it into two traingles of equal areas. 

Therefore, ar(Δ BDF) = ar(Δ DEF).......... (i)

In parallelogram AFDE, 

ar(Δ AFE) = ar(Δ DEF)   (EF is a diagonal)......... (ii)

In parallelogram FDCE, 

ar(Δ CDE) = ar(Δ DEF)   (DE is a diagonal)...........(iii)

From (i), (ii) and (iii)

ar(Δ BDF) = ar(Δ AFE) = ar(Δ CDE) = ar(Δ DEF)..........(iv)

If area of traingles are equal then they are congruent.

Hence, Δ DEF is congruent to triangle Δ BDF = Δ AFE = Δ CDE.

In ΔCAB and ΔADB
Angle B is common and Angle A=Angle D
So the triangles are similar

a=cb/p
Now applying pythagoras theorem in 
Δ ABC
H² = P²+B²
BC²=AC²+AB²

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides

Given,

Area of two similar triangles = 12 cm² , 48 cm² .

Area of bigger triangle = 48 cm²

Area of smaller triangle = 12 cm²

We know that, Ratio of the areas of two similar is equal to square of the ratio of corresponding heights or altitudes.

Given, Height of smaller triangle =2.1 cm.

4 = (Height of the triangle)² / 4.41

√(4 * 4.41) = Height of the bigger triangle

Height of the bigger triangle = 2 * 2.1 = 4.2 cm.

The centroid is the intersection of the three medians while the incentre is the intersection of the three (internal) angle bisectors. In an equilateral triangle, each median is also an angle bisector (and vice versa), the centroid coincides with the incentre. In fact, the centroid, incentre, circumcentre and orthocentre of an equilateral triangle are coincide at the same point.

Construct DG || BF
In triangle ADG,
AE = ED ( given)
EF || DG ( by construction)
a line passing through the mid point of any side of a triangle, parallel to the other side, bisects the third side
So AF = FG    ...(1)
Now in triangle CBF
BD = DC ( given)
DG||BF ( by construction)
a line passing through the mid point of any side of a triangle, parallel to the other side, bisects the third side
So, FG = GC …(2)
Now, by (1) & (2)
AF = FG = GC
 AF : FC = 1:2

since PQ = PR

so ∠Q = ∠R = 40^o

so ∠P comes out to be 100^o

and since PS is a median, so it'll bisect the ∠P into two equal parts

so, ∠QPS = 50^o

Given :△ABC, F, D and E are mid points of AB, BC, CA respectively.
Using mid point theorem we prove that □AEFD, □DBFF and □DCEF are parallelograms. The diagonal of a parallelogram divides the parallelogram into two congruent triangles. So all triangles are congruent to each other. So ΔDEF is congruent to 
△AFE, △BFD and △CDE

We have cosine formula as Cos A = 
So, Cos 60 = 

b²+c²-a²=bc
a2=b2+c² - bc