Question
Solve the following examples.
Do sides 7 cm , 24 cm, 25 cm form a right angled triangle ? Give reason.
Do sides 7 cm , 24 cm, 25 cm form a right angled triangle ? Give reason.
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Use Euclid's division algorithm to find the HCF of:
136, 170 and 255
136, 170 and 255
Solve the following quadratic equation by using formula method:
$4 x^2+x-5=0$
→$4 x^2+x-5=0$
In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F respectively. If AB = 12cm, BC = 8cm and AC = 10cm, find the lengths of AD, BE and CF.
The area enclosed between the concentric circles is $770\ cm^2$^. If the radius of the outer circle is $21\ cm$, find the radius of the inner circle.
→Solve the following quadratic equation by using formula method:
$x^2+2 x-7=0$
→$x^2+2 x-7=0$
$ABC$ is an isosceles triangle, right-angled at $B$. Similar triangle $ACD$ and $ABE$ are constructed on sides $AC$ and $AB$. Find the ratio between the areas of $\triangle\text{ABE}$ and $\triangle\text{ACD}.$
→In figure, chord EF || chord GH. Prove that, chord EG ≅ chord FH. Fill in the blanks and write the proof.
Proof: Draw seg GF.
$\angle EFG =\angle FGH \quad \ldots . . . \square \quad \ldots . . .( I )$
$\angle EFG =\square \quad \ldots . .[\text { [inscribed angle theorem] (II) }$
$\angle FGH =\square \quad \ldots . .[\text { inscribed angle theorem] (III) }$
$\therefore m (\operatorname{arc} EG )=\square \quad \ldots \ldots[ By ( I ),( II ), \text { and (III)] }$
chord $EG \cong$ chord $FH \quad$.........[corresponding chords of congruent arcs]
→Proof: Draw seg GF.
$\angle EFG =\angle FGH \quad \ldots . . . \square \quad \ldots . . .( I )$
$\angle EFG =\square \quad \ldots . .[\text { [inscribed angle theorem] (II) }$
$\angle FGH =\square \quad \ldots . .[\text { inscribed angle theorem] (III) }$
$\therefore m (\operatorname{arc} EG )=\square \quad \ldots \ldots[ By ( I ),( II ), \text { and (III)] }$
chord $EG \cong$ chord $FH \quad$.........[corresponding chords of congruent arcs]
Prove that:
$\frac{\sec(90^\circ-\theta)\text{cosec }\theta-\tan(90^\circ-\theta)\cot\theta+\cos^225^\circ+\cos^265^\circ}{3\tan27^\circ\tan63^\circ}=\frac{2}{3}$
→$\frac{\sec(90^\circ-\theta)\text{cosec }\theta-\tan(90^\circ-\theta)\cot\theta+\cos^225^\circ+\cos^265^\circ}{3\tan27^\circ\tan63^\circ}=\frac{2}{3}$
Prove that:$\tan20^\circ\tan35^\circ\tan45^\circ\tan55^\circ\tan70^\circ=1$
→Find the area of a ring whose outer and inner radii are respectively 23cm and 12cm.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$
→