ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that
(i) ∆ ABE ≅ ∆ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
ABC and DBC are two isosceles triangles on the same base BC (see Fig). Show that ∠ABD = ∠ACD.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠ BCD is a right angle.
Length (in mm)
Number of leaves
118 − 126
3
127 − 135
5
136 − 144
9
145 − 153
12
154 − 162
163 − 171
4
172 − 180
2
(i) Draw a histogram to represent the given data. (ii) Is there any other suitable graphical representation for the same data? (iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.
Length (in hours)
Number of lamps
300 − 400
14
400 − 500
56
500 − 600
60
600 − 700
86
700 − 800
74
800 − 900
62
900 − 1000
48
(i) Represent the given information with the help of a histogram. (ii) How many lamps have a lifetime of more than 700 hours?
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆ PQR (see Fig. 7.40). Show that:
(i) ∆ BM≅∆ PQN
(ii) ∆ ABC≅∆ PQR
Simplify : (i) 2 \(\frac{2}{3}\) . 2 \(\frac{1}{5}\) (ii) (\(\frac{1}{33}\))7 (iii) 11 \(\frac{1}{2}\) / 11\(\frac{1}{4}\) (iv) 7 \(\frac{1}{2}\) . 8 \(\frac{1}{2}\)
Section A
Section B
Marks
Frequency
0 − 10
10 − 20
19
20 − 30
17
15
30 − 40
10
40 − 50
1
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Number of balls
Team A
Team B
1 − 6
7 − 12
6
13 − 18
8
19 − 24
25 − 30
31 − 36
37 − 42
43 − 48
49 − 54
55 − 60
Represent the data of both the teams on the same graph by frequency polygons.[Hint: First make the class intervals continuous.]