Pythagoras Theorem Class 9 ICSE Important Questions with Answers

Answer: (a) $10$10 cm.

Hypotenuse $=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$=√(6²+8²)=√(36+64)=√100=10 cm.

Answer: (b) $(5,12,13)$(5,12,13).

$5^2+12^2=25+144=169=13^2$5²+12²=25+144=169=13², so it satisfies $a^2+b^2=c^2$a²+b²=c². The others do not.

Answer: (b) $8$8 cm.

For a square of side $a$a, diagonal $=a\sqrt2$=a√2. So $a\sqrt2=8\sqrt2$a√2=8√2, giving $a=8$a=8 cm.

Answer: (a) $6$6 cm.

When the altitude is drawn from the right angle to the hypotenuse, $BD^2=AD\times DC=4\times9=36$BD²=AD× DC=4×9=36, so $BD=6$BD=6 cm.

Answer: (a) $9^2+12^2=81+144=225=15^2$9²+12²=81+144=225=15². By the converse of Pythagoras theorem the triangle is right-angled (right angle opposite the $15$15 cm side). R correctly explains A.

The wall, ground and ladder form a right-angled triangle with the ladder as the hypotenuse.

Let the height reached be $h$h. By Pythagoras theorem:

$h^2=13^2-5^2=169-25=144$h²=13²-5²=169-25=144

$h=\sqrt{144}=12\text{ m}.$h=√144=12 m.

The ladder reaches $12$12 m up the wall.

The longest side is $25$25 cm. Check the converse of Pythagoras theorem.

$7^2+24^2=49+576=625$7²+24²=49+576=625.

$25^2=625$25²=625.

Since $7^2+24^2=25^2$7²+24²=25², by the converse of Pythagoras theorem the triangle is right-angled, with the right angle opposite the $25$25 cm side.

The altitude $AD$AD from the apex of an isosceles triangle bisects the base $BC$BC.

So $BD=DC=\dfrac{10}{2}=5$BD=DC=10/2=5 cm, and $\angle ADB=90^\circ$∠ ADB=90°.

In right $\triangle ADB$△ ADB, by Pythagoras theorem:

$AD^2=AB^2-BD^2=13^2-5^2=169-25=144$AD²=AB²-BD²=13²-5²=169-25=144

$AD=\sqrt{144}=12\text{ cm}.$AD=√144=12 cm.

The altitude $AD=12$AD=12 cm.

The eastward and northward paths are perpendicular, forming a right-angled triangle whose hypotenuse is the required distance.

Let the distance from the start be $AC$AC. By Pythagoras theorem:

$AC^2=15^2+8^2=225+64=289$AC²=15²+8²=225+64=289

$AC=\sqrt{289}=17\text{ m}.$AC=√289=17 m.

He is $17$17 m from the starting point.

First find the diagonal $AC$AC in right $\triangle ABC$△ ABC ($\angle B=90^\circ$∠ B=90°):

$AC^2=AB^2+BC^2=3^2+4^2=9+16=25$AC²=AB²+BC²=3²+4²=9+16=25

$AC=5\text{ cm}.$AC=5 cm.

Now in $\triangle ACD$△ ACD, $\angle ACD=90^\circ$∠ ACD=90°, so $AD$AD is the hypotenuse:

$AD^2=AC^2+CD^2=5^2+12^2=25+144=169$AD²=AC²+CD²=5²+12²=25+144=169

$AD=\sqrt{169}=13\text{ cm}.$AD=√169=13 cm.

Therefore $AD=13$AD=13 cm.

The standing part ($5$5 m) is vertical, the distance on the ground ($12$12 m) is horizontal, and the broken part is the hypotenuse.

Broken part $=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$=√(5²+12²)=√(25+144)=√169=13 m.

Original height $=$= standing part $+$+ broken part $=5+13=18$=5+13=18 m.

Given: A right-angled $\triangle ABC$△ ABC with $\angle B=90^\circ$∠ B=90°.

To prove: $AC^2=AB^2+BC^2$AC²=AB²+BC².

Construction: Draw $BD\perp AC$BD⊥ AC, with $D$D on $AC$AC.

Proof: In $\triangle ADB$△ ADB and $\triangle ABC$△ ABC:

$\angle A=\angle A$∠ A=∠ A (common) and $\angle ADB=\angle ABC=90^\circ$∠ ADB=∠ ABC=90°.

By AA similarity, $\triangle ADB\sim\triangle ABC$△ ADB∼ △ ABC, so $\dfrac{AD}{AB}=\dfrac{AB}{AC}$AD/AB=AB/AC, giving

$AB^2=AD\cdot AC \quad\text{...(1)}$AB²=AD· AC ...(1)

Similarly, in $\triangle BDC$△ BDC and $\triangle ABC$△ ABC: $\angle C=\angle C$∠ C=∠ C (common) and $\angle BDC=\angle ABC=90^\circ$∠ BDC=∠ ABC=90°.

By AA similarity, $\triangle BDC\sim\triangle ABC$△ BDC∼ △ ABC, so $\dfrac{DC}{BC}=\dfrac{BC}{AC}$DC/BC=BC/AC, giving

$BC^2=DC\cdot AC \quad\text{...(2)}$BC²=DC· AC ...(2)

Adding (1) and (2):

$AB^2+BC^2=AD\cdot AC+DC\cdot AC=AC(AD+DC)$AB²+BC²=AD· AC+DC· AC=AC(AD+DC)

Since $AD+DC=AC$AD+DC=AC,

$AB^2+BC^2=AC\cdot AC=AC^2.$AB²+BC²=AC· AC=AC².

Hence $AC^2=AB^2+BC^2$AC²=AB²+BC². Proved.

Since $\angle A=90^\circ$∠ A=90°, use Pythagoras theorem in the right triangles.

Let $AB=c$AB=c and $AC=b$AC=b. Then $BC^2=b^2+c^2$BC²=b²+c².

$P$P is mid-point of $AB$AB: $AP=\dfrac{c}{2}$AP=c/2. In right $\triangle APC$△ APC:
$CP^2=AC^2+AP^2=b^2+\dfrac{c^2}{4}.$CP²=AC²+AP²=b²+c²/4.

$Q$Q is mid-point of $AC$AC: $AQ=\dfrac{b}{2}$AQ=b/2. In right $\triangle ABQ$△ ABQ:
$BQ^2=AB^2+AQ^2=c^2+\dfrac{b^2}{4}.$BQ²=AB²+AQ²=c²+b²/4.

Adding:
$BQ^2+CP^2=c^2+\dfrac{b^2}{4}+b^2+\dfrac{c^2}{4}=\dfrac54(b^2+c^2).$BQ²+CP²=c²+b²/4+b²+c²/4=5/4(b²+c²).

Multiply by $4$4:
$4(BQ^2+CP^2)=5(b^2+c^2)=5BC^2.$4(BQ²+CP²)=5(b²+c²)=5BC².

Hence proved.

(i) In right $\triangle PQR$△ PQR ($\angle Q=90^\circ$∠ Q=90°):
$PR^2=PQ^2+QR^2=40^2+30^2=1600+900=2500$PR²=PQ²+QR²=40²+30²=1600+900=2500
$PR=\sqrt{2500}=50\text{ m}.$PR=√2500=50 m.

(ii) The pole ($24$24 m) is vertical at $P$P and $PR=50$PR=50 m is horizontal. The wire is the hypotenuse:
$\text{wire}=\sqrt{24^2+50^2}=\sqrt{576+2500}=\sqrt{3076}\approx55.5\text{ m}.$wire=√(24²+50²)=√(576+2500)=√3076≈55.5 m.

(iii) Route $PQ+QR=40+30=70$PQ+QR=40+30=70 m; diagonal $PR=50$PR=50 m. The diagonal is $70-50=20$70-50=20 m shorter.

(iv) Pythagoras theorem: in a right triangle, hypotenuse$^2=$²= sum of squares of the other two sides; its converse: if one side's square equals the sum of the squares of the other two, the triangle is right-angled.