CBSE Class 10 Mathematics Triangles Worksheet Set 01

Case Study Based Questions

I. Rahul is studying in X Standard. He is making a kite to fly it on a Sunday. Few questions came to his mind while making the kite. Give answers to his questions by looking at the figure.

Question. Rahul tied the sticks at what angles to each other?
(a) 30°
(b) 45°
(c) 90°
(d) 60°
Answer : C

Question. Which is the correct similarity criteria applicable for smaller triangles at the upper part of this kite?
(a) RHS
(b) SAS
(c) SSA
(d) AAS
Answer : B

Question. Sides of two similar triangles are in the ratio 4:9. Corresponding medians of these triangles are in the ratio
(a) 2:3
(b) 4:9
(c) 81:16
(d) 16:81
Answer : B

Question. In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. This theorem is called as,
(a) Pythagoras theorem
(b) Thales theorem
(c) Converse of Thales theorem
(d) Converse of Pythagoras theorem
Answer : D

Question. What is the area of the kite, formed by two perpendicular sticks of length 6 cm and 8 cm?
(a) 48 cm²
(b) 14 cm²
(c) 24 cm²
(d) 96 cm²
Answer : A
 

II. For going to city B from city A, there is a route via city C such that AC ⊥ CB, AC = 2x km and CB = 2(x + 7) km. It is proposed to construct a 26 km highway which is directly connect the two cities A and B. In solving such problem authority of the cities have some questions in his mind. Give answer to his following questions:

Question. Which theorem will you use to solve this problem?
(a) Pythagoras theorem
(b) Basic proportionality theorem
(c) Factor theorem
(d) Fundamental Theorem of Arithmetic
Answer : A

Question. What is the distance of AC?
(a) 15 km
(b) 18 km
(c) 10 km
(d) 20 km
Answer : C

Question. What is the distance of BC?
(a) 12 km
(b) 10 km
(c) 20 km
(d) 24 km
Answer : A

Question. Find how much distance will be saved in reaching city B from city A after the construction of the highway.
(a) 6 km
(b) 8 km
(c) 10 km
(d) 12 km
Answer : B

Question. If ∠A is supposed to be 30°, then ∠B = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer : C
 

III. Ravi is a student of X^th standard. He is drawing a square on a sheet of paper considering each side of the square as 8 cm. Now, he is drawing two equilateral triangles, one by taking side of the square as one of its side and another by taking diagonal of square as sides of another equilateral triangle. Later, few questions came to his mind and he wants to solve them. Give answers to his questions by drawing your own figure.

Question. The diagonal of the square is
(a) 8√2 cm
(b) 8√3 cm
(c) 8√5 cm
(d) 8√6 cm
Answer : A

Question. The area of the square is
(a) 64 sq. cm
(b) 81 sq. cm
(c) 121 sq. cm
(d) 49 sq. cm
Answer : A

Question. The area of equilateral triangle by taking side of square as side of the triangle is
(a) 8√3 sq. cm
(b) 16√3 sq. cm
(c) 32√3 sq. cm
(d) 24√3 sq. cm
Answer : B

Question. The area of equilateral triangle by taking diagonal of square as the side of the equilateral triangle is
(a) 8√3 sq. cm
(b) 16√3 sq. cm
(c) 32√3 sq. cm
(d) 64√3 sq. cm
Answer : C

Question. Which is true order in area?
(a) Equilateral triangle on side < Square < Equilateral triangle with diagonal
(b) Equilateral triangle on side > Square > Equilateral triangle on diagonal
(c) Equilateral triangle on diagonal > Equilateral triangle on side > Square
(d) Square > Equilateral triangle on diagonal > Equilateral triangle on side
Answer : D

 

Again, EF || AB || DC

=> FP || AB

Thus, in ΔCAB, we have

FP || BA

Therefore, by basic proportionality theorem,

we have   BF   =   AP     ....(ii)

               FC        PC

From (i) and (ii), we have   AE   =   BF

                                        ED        FC

 

Question. In figure, ∠A = ∠ B and DE || BC. Prove that AD = BE

 

 

Question. In fig., DE || BC. If AD = 4x – 3, DB = 3x – 1,AE = 8x – 7 and EC = 5x – 3, find the value of x.

.

Question 1. OBJECTIVE: To illustrate that the medians of a triangle are concurrent and the point of concurrency (centroid) always lies inside the triangle.
Answer: To achieve this objective, we follow the step-by-step activity procedure using paper folding:
1. Knowledge of different types of triangles.
2. Concept of median line segment.
3. Concept of concurrent lines.
4. Knowledge of obtaining the mid-point of a line segment by paper folding.

Procedure:
1. Take a coloured paper and cut an acute angled triangle \(ABC\) from this paper.
2. Fold the side \(BC\) of \(\triangle ABC\) keeping the point \(B\) exactly on the point \(C\) and press the fold such that the mid-point \(D\) of \(BC\) is obtained.
3. Similarly, repeat this process to find the mid-points \(E\) and \(F\) of other two sides \(AC\) and \(AB\) respectively.
4. Fold the \(\triangle ABC\) along \(AD\) and press the paper to get a clear crease.
5. Mark a line by a pen on this crease \(AD\) as a median of \(\triangle ABC\).
6. Similarly, obtain the other two medians \(BE\) and \(CF\).
7. Repeat the same activity for a right angled and an obtuse angled triangle.

Observation:
1. We observe that in each case, all the 3 medians pass through the same point \(G\) which shows that the medians of a triangle are concurrent.
2. The point of concurrency of the medians (the centroid) of the triangle always lies inside the triangle.
In simple words: This activity proves that when you draw lines from all three corners of any triangle to the middle of the opposite sides, all three lines always meet at a single interior point called the centroid.

Exam Tip: Remember that the point where all three medians intersect is called the centroid, and it divides each median in the ratio \(2:1\). It is always located inside the triangle regardless of whether the triangle is acute, right, or obtuse.