Chapter 6 – Triangles
Diagonals of a trapezium ABCD with AB∥DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD
05
Oct
Diagonals of a trapezium ABCD with AB∥DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD Diagonals of a trapezium ABCD with AB∥DC intersect each other at the point O. If AB = 2 CD find the ratio of the areas [...]
Let triangle abc similar to triangle def and their areas be, respectively 64 cm^2 and 121 cm^2 . If ef = 15.4 cm, find bc.
05
Oct
Let triangle abc similar to triangle def and their areas be, respectively 64 cm^2 and 121 cm^2 . If ef = 15.4 cm, find bc. find bc. Let triangle abc similar to triangle def and their areas be respectively 64 cm^2 and 121 cm^2 . If ef = 15.4 cm October 5, 2020 Category: Chapter [...]
If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ∼ ΔPQR prove that AB/PQ = AD/PM.
05
Oct
If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ∼ ΔPQR prove that AB/PQ = AD/PM. If AD and PM are medians of triangles ABC and PQR respectively where ΔABC ∼ ΔPQR prove that AB/PQ = AD/PM. October 5, 2020 Category: Chapter 6 - Triangles , Maths , NCERT Class [...]
A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
05
Oct
A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same [...]
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ∼ ΔPQR.
05
Oct
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ∼ ΔPQR. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another [...]
D is a point on the side BC of a triangle ABC such that angle ADC = angle BAC Show that CA square = CB.CD.
05
Oct
D is a point on the side BC of a triangle ABC such that angle ADC = angle BAC Show that CA square = CB.CD. D is a point on the side BC of a triangle ABC such that angle ADC = angle BAC Show that CA square = CB.CD. October 5, 2020 Category: Chapter [...]
Sides AB and BC and Median AD of a triangle ABC are proportional to sides PQ and PR and median PM of triangle PQR. Show that triangle ABC ~ triangle PQR .
05
Oct
Sides AB and BC and Median AD of a triangle ABC are proportional to sides PQ and PR and median PM of triangle PQR. Show that triangle ABC ~ triangle PQR . Sides AB and BC and Median AD of a triangle ABC are proportional to sides PQ and PR and median PM of triangle [...]
In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ~ Δ ECF.
05
Oct
In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ~ Δ ECF. E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ [...]
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ∼ ΔFEG, Show that: (i) CD/GH = AC/FG (ii) ΔDCB ∼ ΔHGE (iii) ΔDCA ∼ ΔHGF
05
Oct
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ∼ ΔFEG, Show that: (i) CD/GH = AC/FG (ii) ΔDCB ∼ ΔHGE (iii) ΔDCA ∼ ΔHGF CD and GH are respectively the bisectors of ∠ACB and [...]
In Fig., ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) △ABC∼△AMP (ii) PA/CA = MP/BC.
04
Oct
In Fig., ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) △ABC∼△AMP (ii) PA/CA = MP/BC. ABC and AMP are two right triangles In Fig. right angled at B and M respectively. Prove that: (i) △ABC∼△AMP (ii) PA/CA = MP/BC. October 4, 2020 Category: Chapter 6 - [...]